Collisions.

Collisions

Introduction In this chapter you will learn to solve problems involving the collision of 2 or more particles You will learn about the coefficient of restitution between particles and surfaces You will use vector notation and the formulae for conservation of momentum from M1 You will also see a lot of algebraic examples rather than just numerical ones!

Teachings for Exercise 4A

Collisions 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can apply the impulse-momentum principle and the principle of conservation of linear momentum in vector form You have already seen the following formulae in M1: I = impulse m = mass v = final velocity u = initial velocity t = time The 1 and 2 are used to represent 2 different particles. Impulse is measured is Ns Momentum is measured in kgms-1 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 4A

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can apply the impulse-momentum principle and the principle of conservation of linear momentum in vector form A particle of mass 0.2kg is moving with velocity (10i - 5j)ms-1 when it receives an impulse of (3i – 2j)Ns. Find the new velocity of the particle. 𝑰=𝑚𝒗−𝑚𝒖 Sub in values. We need to find the final velocity, v 3𝒊−2𝒋=0.2𝒗−0.2(10𝒊−5𝒋) Multiply out the bracket 3𝒊−2𝒋=0.2𝒗−2𝒊+𝒋 Add 2i, subtract j 5𝒊−3𝒋=0.2𝒗 Multiply by 5 25𝒊−15𝒋=𝒗 4A

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 You can apply the impulse-momentum principle and the principle of conservation of linear momentum in vector form An ice hockey puck of mass 0.17kg receives an impulse of QNs. Immediately before the impulse the velocity of the puck is (10i + 5j)ms-1 and immediately afterwards its velocity is (15i – 7j)ms-1. Find the magnitude of Q and the angle between Q and i. Sub in values 𝑰= 𝒊−7𝒋 −0.17(10𝒊+5𝒋) Multiply out brackets 𝑰=2.55𝒊−1.19𝒋−1.7𝒊−0.85𝒋 Simplify terms 𝑰=0.85𝒊−2.04𝒋 Use Pythagoras’ Theorem to find the magnitude of the impulse Q 0.85i 𝑰 = (−2.04) 2 Calculate 𝑰 =2.21 𝑰 = -2.04j 0.85i – 2.04j Calculate 𝑰 =2.21 Q 4A

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 You can apply the impulse-momentum principle and the principle of conservation of linear momentum in vector form An ice hockey puck of mass 0.17kg receives an impulse of QNs. Immediately before the impulse the velocity of the puck is (10i + 5j)ms-1 and immediately afterwards its velocity is (15i – 7j)ms-1. Find the magnitude of Q and the angle between Q and i. Sub in values 𝑰= 𝒊−7𝒋 −0.17(10𝒊+5𝒋) Multiply out brackets 𝑰=2.55𝒊−1.19𝒋−1.7𝒊−0.85𝒋 Simplify terms 𝑰=0.85𝒊−2.04𝒋 To find the angle, consider the direction of the vector Q and the direction of i. Adj 𝜃=𝑇𝑎 𝑛 −1 𝑂 𝐴 0.85i i Sub in values θ 𝑰 =2.21 𝜃=𝑇𝑎 𝑛 − -2.04j 𝜃=67.4° 0.85i – 2.04j Opp Calculate θ Q 𝜃=67.4° 4A

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can apply the impulse-momentum principle and the principle of conservation of linear momentum in vector form A squash ball of mass 0.025kg is moving with velocity (22i + 37j)ms-1 when it hits a wall. It rebounds with velocity (10i – 11j)ms-1. Find the impulse exerted by the wall on the squash ball.  The impulse exerted by the wall on the squash ball is equal to the change in momentum of the ball. 𝑰=𝑚𝒗−𝑚𝒖 Sub in values 𝑰= 𝒊−11𝒋 −0.025(22𝒊+37𝒋) Multiply out brackets 𝑰=0.25𝒊−0.275𝒋−0.55𝒊−0.925𝒋 Simplify 𝑰= −0.3𝒊−1.2𝒋 𝑁𝑠 4A

Collisions 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can apply the impulse-momentum principle and the principle of conservation of linear momentum in vector form A particle of mass 0.15kg is moving with velocity (20i – 10j)ms-1 when it collides with a particle of mass 0.25kg moving with velocity (16i – 8j)ms-1. The two particles coalesce and form one particle of mass 0.4kg.  ‘Coalesce’ means ‘join together’ Find the velocity of the combined particle. Use the formula for the conservation of momentum  As the particles combine, the right hand side will share a common mass and velocity Sub in values 𝒊−10𝒋 𝒊−8𝒋 =0.4𝒗 Multiply out brackets 3𝒊−1.5𝒋+4𝒊−2𝒋=0.4𝒗 Simplify 7𝒊−3.5𝒋=0.4𝒗 Divide by 0.4 17.5𝒊−8.75𝒋=𝒗 4A

Teachings for Exercise 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Newton’s law of restitution defines how the speed of the particles after a collision depends on their nature. You can think of restitution as ‘bounciness’ Particles that are more ‘bouncy’ will have a higher coefficient of restitution The coefficient of restitution is calculated using the formula to the right 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 This effectively tells you what fraction of the original speed is maintained after the collision The value e is the coefficient of restitution and: 0≤𝑒≤1 Perfectly elastic particles will have an e value of 1 and perfectly inelastic particles will have a value of 0 To give you a rough idea of some objects.. A Table tennis ball has a value of around 0.95 Tennis, golf and cricket balls range from 0.4 to 0.9 A ball of plasticine will have a value very close to 0 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Perfectly elastic particles You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Newton’s law of restitution defines how the speed of the particles after a collision depends on their nature. You can think of restitution as ‘bounciness’ Particles that are more ‘bouncy’ will have a higher coefficient of restitution The coefficient of restitution is calculated using the formula to the right v v Before collision: v v After collision: Perfectly inelastic particles v v Before collision: 0ms-1 After collision: 4B

Applet for collision demonstrations
𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution In these questions the diagrams show the speeds of two particles A and B just before and just after a collision. The particles are moving on a smooth horizontal plane. Find the coefficient of restitution in each case. (You should always set up diagrams like these, especially if you aren’t given them before hand) Before impact After impact 8 2 At rest At rest A B A B 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 The speed of approach You need to find the difference in the speeds of approach Speed of A – Speed of B 8 – 0 = 8 The speed of separation You need to find the difference in the speeds of separation This needs to be done the opposite way as B is now moving away from A Speed of B – Speed of A 2 – 0 = 2 Speed of approach = 8 Speed of separation = 2 Whichever way you perform the first subtraction, the second must be done the opposite way! 4B

𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution In these questions the diagrams show the speeds of two particles A and B just before and just after a collision. The particles are moving on a smooth horizontal plane. Find the coefficient of restitution in each case. (You should always set up diagrams like these, especially if you aren’t given them before hand) Before impact After impact 8 2 At rest At rest A B A B 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 𝑒= 2 8 Simplify 𝑒= 1 4 Speed of approach = 8 Speed of separation = 2 4B

𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution In these questions the diagrams show the speeds of two particles A and B just before and just after a collision. The particles are moving on a smooth horizontal plane. Find the coefficient of restitution in each case. (You should always set up diagrams like these, especially if you aren’t given them before hand) Before impact After impact 6 3 4 5 A B A B Approach 6 – 3 = 3 Separation 5 – 4 = 1 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 𝑒= 1 3 4B

𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution In these questions the diagrams show the speeds of two particles A and B just before and just after a collision. The particles are moving on a smooth horizontal plane. Find the coefficient of restitution in each case. (You should always set up diagrams like these, especially if you aren’t given them before hand) Before impact After impact 11 7 6 3 A B A B Approach = 18 Separation = 9 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 𝑒= 9 18 Simplify 𝑒= 1 2 Remember to use negative numbers if particles are travelling in opposite directions! 4B

Collisions 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 4B
𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Find the value of v in the situation shown, given that e = 1/3 Before impact After impact 4 3 2 v A B A B Approach 4 – 3 = 1 Separation v - 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values, leave the speed of separation in algebraic form 1 3 = 𝑣−2 1 Multiply by 1! 1 3 =𝑣−2 Add 2 2 1 3 =𝑣 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution You can use the principle of conservation of linear momentum together with Newton’s Law of Restitution to solve problems involving two unknown velocities In the example shown, calculate the values of v1 and v2, given that the coefficient of restitution is 1/2  You will need to use each of the above rules to form two equations, which you can then solve simultaneously 5 4 v1 v2 A B A B 200g 400g 200g 400g Approach = 9 Separation v2 – v1 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 1 2 = 𝑣 2 − 𝑣 1 9 Multiply by 9 4.5= 𝑣 2 − 𝑣 1 Double all (to remove decimals) 9= 2𝑣 2 −2 𝑣 1 9= 2𝑣 2 −2 𝑣 1 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution You can use the principle of conservation of linear momentum together with Newton’s Law of Restitution to solve problems involving two unknown velocities In the example shown, calculate the values of v1 and v2, given that the coefficient of restitution is 1/2  You will need to use each of the above rules to form two equations, which you can then solve simultaneously 5 4 v1 v2 A B A B 200g 400g 200g 400g 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values (0.2)(5)+(0.4)(−4)=(0.2)( 𝒗 1 )+(0.4)( 𝒗 2 ) Work out terms 1−1.6=0.2 𝑣 𝑣 2 Simplify left side −0.6=0.2 𝑣 𝑣 2 Multiply by 10 (to remove the decimal) −6=2 𝑣 1 +4 𝑣 2 Divide by 2 (to simplify) −3= 𝑣 1 +2 𝑣 2 9= 2𝑣 2 −2 𝑣 1 −3= 𝑣 1 +2 𝑣 2 4B

Collisions 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑣 1 =−4
𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution You can use the principle of conservation of linear momentum together with Newton’s Law of Restitution to solve problems involving two unknown velocities In the example shown, calculate the values of v1 and v2, given that the coefficient of restitution is 1/2  You will need to use each of the above rules to form two equations, which you can then solve simultaneously 𝑣 1 =−4 5 4 v1 4 1/2 v2 𝑣 2 = 1 2 A B A B 200g 400g 200g 400g 1) 9= 2𝑣 2 −2 𝑣 1 Eliminate v2 by subtracting 1 from 2 (be careful with negatives) 2) −3= 𝑣 1 +2 𝑣 2 2) – 1) −12= 3𝑣 1 Divide by 3 −4= 𝑣 1 Now sub this into one of the equations to find v2 2) −3= 𝑣 1 +2 𝑣 2 Sub in v1 −3=−4+2 𝑣 2 Add 4 1=2 𝑣 2 9= 2𝑣 2 −2 𝑣 1 Divide by 2 1 2 = 𝑣 2 −3= 𝑣 1 +2 𝑣 2 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 5𝑢𝑒= 𝑣 2 − 𝑣 1 3u 2u v1 v2 P Q P Q 3m 4m 3m 4m Approach 3u - - 2u = 5u Separation v2 – v1 Speed of Q after the collision = v2  We need to set up simultaneous equations and solve then for v2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 𝑒= 𝑣 2 − 𝑣 1 5𝑢 Multiply by 5u 5𝑢𝑒= 𝑣 2 − 𝑣 1 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 5𝑢𝑒= 𝑣 2 − 𝑣 1 3u 2u v1 v2 𝑢=3 𝒗 1 +4 𝒗 2 P Q P Q 3m 4m 3m 4m 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values 3𝑚 3𝑢 + 4𝑚 (−2𝑢)=(3𝑚)( 𝒗 1 )+(4𝑚)( 𝒗 2 ) Work out terms 9𝑚𝑢−8𝑚𝑢=3𝑚 𝒗 1 +4𝑚 𝒗 2 Cancel m’s 9𝑢−8𝑢=3 𝒗 1 +4 𝒗 2 Simplify 𝑢=3 𝒗 1 +4 𝒗 2 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 5𝑢𝑒= 𝑣 2 − 𝑣 1 3u 2u v1 v2 𝑢=3 𝒗 1 +4 𝒗 2 P Q P Q 3m 4m 3m 4m Rewrite in terms of v1 1) 5𝑢𝑒= 𝑣 2 − 𝑣 1 1) 𝑣 1 = 𝑣 2 −5𝑢𝑒 2) 𝑢=3 𝒗 1 +4 𝒗 2 𝑢=3 𝒗 1 +4 𝒗 2 Replace v1 𝑢=3( 𝑣 2 −5𝑢𝑒)+4 𝒗 2 Multiply out 𝑢=3 𝑣 2 −15𝑢𝑒+4 𝒗 2 Simplify 𝑢=7 𝑣 2 −15𝑢𝑒 Add 15ue 15𝑢𝑒+𝑢=7 𝑣 2 Factorise left side 𝑢(15𝑒+1)=7 𝑣 2 Divide by 7 𝑢 7 (15𝑒+1)= 𝑣 2 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 5𝑢𝑒= 𝑣 2 − 𝑣 1 3u 2u v1 v2 𝑢=3 𝒗 1 +4 𝒗 2 P Q P Q 3m 4m 3m 4m Part b) refers to the new speed of P. We will therefore have to calculate v1 in the same way we found v2. Rewrite in terms of v2 1) 5𝑢𝑒= 𝑣 2 − 𝑣 1 5𝑢𝑒+ 𝑣 1 = 𝑣 2 2) 𝑢=3 𝒗 1 +4 𝒗 2 𝑢=3 𝒗 1 +4 𝒗 2 Replace v2 𝑢=3 𝒗 1 +4(5𝑢𝑒+ 𝑣 1 ) Multiply out the bracket 𝑢=3 𝒗 1 +20𝑢𝑒+4 𝑣 1 Group terms 𝑢=7 𝒗 1 +20𝑢𝑒 Subtract 20ue 𝑢−20𝑢𝑒=7 𝒗 1 Factorise the left 𝑢(1−20𝑒)=7 𝒗 1 Divide by 7 𝑢 7 (1−20𝑒)= 𝒗 1 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 5𝑢𝑒= 𝑣 2 − 𝑣 1 3u 2u v1 v2 𝑢=3 𝒗 1 +4 𝒗 2 P Q P Q 3m 4m 3m 4m 𝑢 7 (1−20𝑒)= 𝒗 1 As the direction of motion of P is unchanged, v1 > 0 As u/7 is greater than 0, The expression in the bracket must also be greater than 0… 1−20𝑒>0 Add 20e 𝑒< 1 20 1>20𝑒 Divide by 20 1 20 >𝑒 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact 𝑣 1 = 𝑢 7 (1−20𝑒) You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 3u 2u v1 v2 𝑣 2 = 𝑢 7 (15𝑒+1) P Q P Q 3m 4m 3m 4m Impulse of P on Q = change in momentum of Q 𝑰=𝑚𝒗−𝑚𝒖 Sub in values 𝑰=(4𝑚) 𝑢 7 15𝑒+1 −(4𝑚)(−2𝑢) Multiply out a bracket 𝑰= 4𝑚𝑢 7 15𝑒+1 −(−8𝑚𝑢) Simplify 𝑰= 60𝑚𝑢𝑒 7 + 4𝑚𝑢 7 +8𝑚𝑢 4B

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Applet for collision demonstrations Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Before impact After impact 𝑣 1 = 𝑢 7 (1−20𝑒) You can solve problems involving the direct impact of two particles by using conservation of linear momentum and Newton’s Law of Restitution Two small spheres have mass 3m and 4m respectively. They are moving towards each other in opposite directions on a smooth horizontal plane. P has speed 3u and Q has speed 2u just before the impact. The coefficient of restitution between P and Q is e. Show that the speed of Q after the collisions is given by u/7(15e + 1) Given that the direction of motion of P is unchanged, find the range of possible values for e Given that the magnitude of the impulse of P on Q is 80mu/9, find the value of e 3u 2u v1 v2 𝑣 2 = 𝑢 7 (15𝑒+1) P Q P Q 3m 4m 3m 4m Impulse of P on Q = change in momentum of Q 𝑰= 60𝑚𝑢𝑒 7 + 4𝑚𝑢 7 +8𝑚𝑢 Sub in the value we are given for the impulse 80𝑚𝑢 9 = 60𝑚𝑢𝑒 7 + 4𝑚𝑢 7 +8𝑚𝑢 Divide by mu 80 9 = 60𝑒 Make the fractions equivalent = 540𝑒 Multiply all terms by 63 560=540𝑒 Subtract 504 and subtract 36 20=540𝑒 1 27 =𝑒 Divide by 540 4B

Teachings for Exercise 4C

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 Before impact After impact You can also apply Newton’s Law of Restitution to problems involving direct collision with a smooth plane surface perpendicular to the direction of motion (ie – a wall!) The diagram shows the motion of an object bouncing off a smooth plane surface u v Approach u – 0 = u Separation 0 - - v = v The smooth plane can be thought of as having an initial speed and final speed of 0 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 𝑒= 𝑣 𝑢 You can use this formula for the coefficient of restitution for a particle colliding with a perpendicular plane 4C

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 Before impact After impact You can also apply Newton’s Law of Restitution to problems involving direct collision with a smooth plane surface perpendicular to the direction of motion (ie – a wall!) A particle collides normally with a fixed vertical plane. The diagram shows the speeds (in ms-1) of the particle before and after collision. Find the value of the coefficient of restitution, e. 8 2 𝑒= 𝑣 𝑢 Sub in values 𝑒= 2 8 Simplify 𝑒= 1 4 4C

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑒= 𝑣 𝑢 You can also apply Newton’s Law of Restitution to problems involving direct collision with a smooth plane surface perpendicular to the direction of motion (ie – a wall!) A small sphere collides normally with a fixed vertical wall. Before the impact, the sphere is moving with a speed of 4ms-1 on a smooth horizontal floor. The coefficient of restitution between the sphere and the wall is 0.2. Find the speed of the sphere after the collision. Sub in values 0.2= 𝑣 4 Multiply by 4 0.8=𝑣 4C

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑠=0.225
Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 Remember that we need units in metres! 𝑠=0.225 You can also apply Newton’s Law of Restitution to problems involving direct collision with a smooth plane surface perpendicular to the direction of motion (ie – a wall!) A particle falls 22.5cm from rest onto a smooth horizontal plane. It then rebounds to a height of 10cm. Find the coefficient of restitution between the particle and the plane. Give your answer to 2sf. You will need to find the velocity on impact and after impact To do this, use the SUVAT equations 𝑎=9.8 𝑢=0 22.5cm 𝑡= ? 𝑣= ? Finding the velocity on impact 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values 𝑣 2 = (0) 2 + 2(9.8)(0.225) Work out the right side 𝑣 2 =4.41 Square root answer 𝑣=2.1𝑚 𝑠 −1 This is our value for u, the initial speed of the particle before colliding with the plane 𝑢=2.1𝑚 𝑠 −1 4C

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 𝑠=0.1
Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 Remember that we need units in metres! 𝑠=0.1 You can also apply Newton’s Law of Restitution to problems involving direct collision with a smooth plane surface perpendicular to the direction of motion (ie – a wall!) A particle falls 22.5cm from rest onto a smooth horizontal plane. It then rebounds to a height of 10cm. Find the coefficient of restitution between the particle and the plane. Give your answer to 2sf. You will need to find the velocity on impact and after impact To do this, use the SUVAT equations 𝑎=−9.8 𝑢= ? 10cm 𝑡= ? 𝑣=0 Finding the velocity after impact  The ball bounces to a height of 10cm, against gravitational acceleration of 9.8. At the height of 10cm, the velocity is 0 We need to find the rebound velocity that will make this happen 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values (0) 2 = 𝑢 2 +2(−9.8)(0.1) Calculate terms 0= 𝑢 2 −1.96 Add 1.96 This is our value for v, the rebound speed of the particle 1.96= 𝑢 2 Square root 1.4=𝑢 𝑢=2.1𝑚 𝑠 −1 𝑣=1.4𝑚 𝑠 −1 4C

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖
Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑭𝑡 𝑰=𝑚𝒗−𝑚𝒖 You can also apply Newton’s Law of Restitution to problems involving direct collision with a smooth plane surface perpendicular to the direction of motion (ie – a wall!) A particle falls 22.5cm from rest onto a smooth horizontal plane. It then rebounds to a height of 10cm. Find the coefficient of restitution between the particle and the plane. Give your answer to 2sf. You will need to find the velocity on impact and after impact To do this, use the SUVAT equations 𝑒= 𝑣 𝑢 Sub in values 𝑒= Simplify 𝑒= 2 3 𝑢=2.1𝑚 𝑠 −1 𝑣=1.4𝑚 𝑠 −1 4C

Teachings for Exercise 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 2=𝑦−𝑥 4D
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations Sphere A colliding with Sphere B You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Before impact After impact 2=𝑦−𝑥 7 3 x y A B A B 1 2 1 2 Approach 7 – 3 = 4 Separation y – x 𝑒= 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ Sub in values 1 2 = 𝑦−𝑥 4 Multiply by 4 2=𝑦−𝑥 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 2=𝑦−𝑥
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations Sphere A colliding with Sphere B You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Before impact After impact 2=𝑦−𝑥 7 3 x y 13=𝑥+2𝑦 A B A B 1 2 1 2 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values = 1 𝑥 +(2)(𝑦) Calculate terms and simplify 13=𝑥+2𝑦 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations Sphere A colliding with Sphere B You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Before impact After impact 2=𝑦−𝑥 7 3 3 x 5 y 13=𝑥+2𝑦 A B A B 1 2 1 2 1) 2=𝑦−𝑥 Add them together to cancel the x terms 2) 13=𝑥+2𝑦 1) + 2) 15=3𝑦 Divide by 3 5=𝑦 Use this to find x 3=𝑥 After the first collision, sphere A is travelling at 3ms-1, sphere B is travelling at 5ms-1 and sphere C is still travelling at 1ms-1 as it has not been affected yet 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 1=𝑏−𝑎 4D
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations New speeds: A = 3ms-1, B = 5ms-1, C = 1ms-1 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Sphere B colliding with Sphere C Before impact After impact 1=𝑏−𝑎 5 1 a b B C B C 2 3 2 3 Approach 5 – 1 = 4 Separation b – a 𝑒= 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ Sub in values 1 4 = 𝑏−1 4 Multiply by 4 1=𝑏−𝑎 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 1=𝑏−𝑎
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations New speeds: A = 3ms-1, B = 5ms-1, C = 1ms-1 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Sphere B colliding with Sphere C Before impact After impact 1=𝑏−𝑎 5 1 a b 13=2𝑎+3𝑏 B C B C 2 3 2 3 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values = 2 𝑎 +(3)(𝑏) Calculate terms and simplify 13=2𝑎+3𝑏 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 1=𝑏−𝑎
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations New speeds: A = 3ms-1, B = 5ms-1, C = 1ms-1 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Sphere B colliding with Sphere C Before impact After impact 1=𝑏−𝑎 5 1 a b 13=2𝑎+3𝑏 B C B C 2 3 2 3 Multiply by 2 1) 1=𝑏−𝑎 3) 2=2𝑏−2𝑎 2) 13=2𝑎+3𝑏 2) 13=2𝑎+3𝑏 Add equations 2 and 3 together to cancel out the ‘a’ terms 3) 2=2𝑏−2𝑎 The final speed of B is 2ms-1 and the final speed of C is 3ms-1 15=5𝑏 Divide by 5 3=𝑏 Use this to find a 2=𝑎 4D

Collisions Speeds after BOTH collisions:
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. Three spheres A, B and C have masses 1kg, 2kg and 3kg respectively. They are moving along the same straight horizontal plane with A following B, which is following C. The initial velocities of A, B and C are 7ms-1, 3ms-1 and 1ms-1 in the direction ABC. Sphere A collides with sphere B and sphere B collides with sphere C. The coefficient of restitution between A and B is 1/2 and between B and C is 1/4 Find the velocities of the 3 spheres after both collisions have taken place b) Explain how you know that there will be a further collision between A and B  Consider each collision separately, drawing diagrams each time. Speeds after BOTH collisions: A = 3ms-1, B = 2ms-1, C = 3ms-1 How do we know there will be a further collision between A and B?  A is travelling faster than B in the same direction, and with no resistances will eventually catch up! 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 3𝑢𝑒=𝑤−𝑣
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations 3𝑢𝑒=𝑤−𝑣 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Find expressions for v and w in terms of u and e. Show that, if the direction of motion of P is changed by the collision, then e > 1/3  Follow the same process, just using algebra instead of numbers Before impact After impact u 2u v w P Q P Q 3m m 3m m Approach u - - 2u = 3u Separation w – v 𝑒= 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ Sub in values 𝑒= 𝑤−𝑣 3𝑢 Multiply by 3u 3𝑢𝑒=𝑤−𝑣 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations 3𝑢𝑒=𝑤−𝑣 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Find expressions for v and w in terms of u and e. Show that, if the direction of motion of P is changed by the collision, then e > 1/3  Follow the same process, just using algebra instead of numbers Before impact After impact u 2u v w 𝑢=3𝑣+𝑤 P Q P Q 3m m 3m m 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values (3𝑚)(𝑢)+(𝑚)(−2𝑢)=(3𝑚)(𝑣)+(𝑚)(𝑤) Simplify 3𝑚𝑢−2𝑚𝑢=3𝑚𝑣+𝑚𝑤 Group terms 𝑚𝑢=3𝑚𝑣+𝑚𝑤 Divide by m 𝑢=3𝑣+𝑤 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations 3𝑢𝑒=𝑤−𝑣 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Find expressions for v and w in terms of u and e. Show that, if the direction of motion of P is changed by the collision, then e > 1/3  Follow the same process, just using algebra instead of numbers Before impact After impact u 2u v w 𝑢=3𝑣+𝑤 P Q P Q 3m m 3m m 1) 3𝑢𝑒=𝑤−𝑣 Rearrange 2) 𝑢=3𝑣+𝑤 𝑢−3𝑣=𝑤 3𝑢𝑒=𝑤−𝑣 Sub the rearrangement in for w 3𝑢𝑒=(𝑢−3𝑣)−𝑣 Group terms 3𝑢𝑒=𝑢−4𝑣 𝒗= 𝒖 𝟒 (𝟏−𝟑𝒆) Add 4v, Subtract 3ue 4𝑣=𝑢−3𝑢𝑒 Factorise right side 4𝑣=𝑢(1−3𝑒) Divide by 4 𝑣= 𝑢 4 (1−3𝑒) 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations 3𝑢𝑒=𝑤−𝑣 You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Find expressions for v and w in terms of u and e. Show that, if the direction of motion of P is changed by the collision, then e > 1/3  Follow the same process, just using algebra instead of numbers Before impact After impact u 2u v w 𝑢=3𝑣+𝑤 P Q P Q 3m m 3m m Rearrange 1) 3𝑢𝑒=𝑤−𝑣 𝑣=𝑤−3𝑢𝑒 2) 𝑢=3𝑣+𝑤 𝑢=3𝑣+𝑤 Replace v with the rearrangement 𝑢=3(𝑤−3𝑢𝑒)+𝑤 Simplify 𝑢=4𝑤−9𝑢𝑒 Add 9ue 𝒗= 𝒖 𝟒 (𝟏−𝟑𝒆) 𝒘= 𝒖 𝟒 (𝟏+𝟗𝒆) 𝑢+9𝑢𝑒=4𝑤 Factorise left side 𝑢(1+9𝑒)=4𝑤 Divide by 4 𝑢 4 (1+9𝑒)=𝑤 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Find expressions for v and w in terms of u and e. Show that, if the direction of motion of P is changed by the collision, then e > 1/3  Follow the same process, just using algebra instead of numbers Before impact After impact u 2u v w P Q P Q 3m m 3m m  If P changes direction, then the final velocity v, must be less than 0 𝑣<0 Sub in the expression for v 𝑢 4 1−3𝑒 <0 u/4 is greater than 0, so the expression in the bracket must be less than 0 1−3𝑒<0 𝒗= 𝒖 𝟒 (𝟏−𝟑𝒆) 𝒘= 𝒖 𝟒 (𝟏+𝟗𝒆) Add 3e 1<3𝑒 Divide by 3 1 3 <𝑒 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Following the collision with P, the sphere Q then collides with and rebounds from a vertical wall. The coefficient of restitution between Q and the wall is e’ c) Given that e = 5/9 and that P and Q collide again in the subsequent motion, show that e’ > 1/9 Before impact After impact 𝒗= 𝒖 𝟒 (𝟏−𝟑𝒆) 𝒗=− 𝒖 𝟔 u 2u v w 𝒘= 𝒖 𝟒 (𝟏+𝟗𝒆) P Q P Q 3m m 3m m  Start by using the value of e to find the values of v and w in terms of u only 𝑣= 𝑢 4 (1−3𝑒) Sub in e 𝑣= 𝑢 4 1−3 5 9 Calculate the part in the bracket 𝑣= 𝑢 4 − 2 3 Calculate v 𝑣=− 𝑢 6 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 𝒗=− 𝒖 𝟔
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Following the collision with P, the sphere Q the collides with and rebounds from a vertical wall. The coefficient of restitution between Q and the wall is e’ c) Given that e = 5/9 and that P and Q collide again in the subsequent motion, show that e’ > 1/9 Before impact After impact 𝒗=− 𝒖 𝟔 u 2u v w 𝒘= 𝒖 𝟒 (𝟏+𝟗𝒆) 𝒘= 𝟑𝒖 𝟐 P Q P Q 3m m 3m m  Start by using the value of e to find the values of v and w in terms of u only 𝑤= 𝑢 4 (1+9𝑒) Sub in e 𝑤= 𝑢 Calculate the part in the bracket 𝑤= 𝑢 4 6 Simplify 𝑤= 3𝑢 2 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Following the collision with P, the sphere Q the collides with and rebounds from a vertical wall. The coefficient of restitution between Q and the wall is e’ c) Given that e = 5/9 and that P and Q collide again in the subsequent motion, show that e’ > 1/9 Before Q hits the wall After Q hits the wall 𝒗=− 𝒖 𝟔 u/6 3u/2 u/6 3ue’/2 P Q P Q 𝒘= 𝟑𝒖 𝟐 3m m 3m m When Q hits the wall it will rebound 𝑒= 𝑣 𝑢 Sub in e = e’, and the approach velocity of Q 𝑒′= 𝑣 3𝑢 2 Multiply by 3u/2 3𝑢𝑒′ 2 =𝑣 This is the velocity which Q will rebound at – we can add this to the diagram above. The impact with the wall has not affected P 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A uniform smooth sphere P of mass 3m is moving in a straight line with speed u on a smooth horizontal table. Another uniform smooth sphere Q of mass m and having the same radius as P, is moving with speed 2u in the opposite direction of P. P and Q collide directly, and their speeds after the collision are v and w respectively. The coefficient of restitution between P and Q is e. Following the collision with P, the sphere Q the collides with and rebounds from a vertical wall. The coefficient of restitution between Q and the wall is e’ c) Given that e = 5/9 and that P and Q collide again in the subsequent motion, show that e’ > 1/9 Before Q hits the wall After Q hits the wall 𝒗=− 𝒖 𝟔 u/6 3u/2 u/6 3ue’/2 P Q P Q 𝒘= 𝟑𝒖 𝟐 3m m 3m m  As there is a further collision, Q must be travelling faster that P in this direction 3𝑢𝑒′ 2 > 𝑢 6 Make denominators equivalent 9𝑢𝑒′ 6 > 𝑢 6 Multiply by 6 9𝑢𝑒′>𝑢 Divide by u 9𝑒′>1 Divide by 9 𝑒′> 1 9 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 4D
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 𝑠=0.9 𝑎=9.8 0.9m 𝑢=0 𝑡= ? 𝑣= ? Approach speed: 4.2ms-1 Height of the first bounce  To start with, we need the approach velocity so we can calculate the speed the ball bounces up from the plane  Use SUVAT 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values 𝑣 2 = (0) 2 +2(9.8)(0.9) Work out terms 𝑣 2 =17.64 Square root 𝑣=4.2𝑚 𝑠 −1 So the ball hits the ground at a velocity of 4.2ms-1 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 𝑠=0.9 𝑎=9.8 0.9m 𝑢=0 𝑡= ? 𝑣= ? Approach speed: 4.2ms-1 Height of the first bounce Rebound speed: 2.1ms-1 𝑒= 𝑣 𝑢 Sub in values to find the rebound velocity 0.5= 𝑣 4.2 Multiply by 4.2 2.1=𝑣 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 𝑠= ? 𝑎=−9.8 0.9m 𝑢=2.1 𝑡= ? ? 𝑣=0 Approach speed: 4.2ms-1 Height of the first bounce  Use the rebound velocity and the fact the velocity at the highest point will be 0 to find the height of the bounce Rebound speed: 2.1ms-1 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values (0) 2 = (2.1) 2 +2(−9.8)𝑠 Work out terms 0=4.41−19.6𝑠 22.5𝑐𝑚 Add 19.6s 19.6𝑠=4.41 Divide by 19.6 𝑠=0.225𝑚 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 𝑠=0 𝑎=−9.8 0.9m 𝑢=2.1 𝑡= ? 0.225m 𝑣= ? Approach speed: 2.1ms-1 Height of the second bounce As before, we need to calculate the velocity that the ball hits the ground with If we consider the first impact with the ground, then s = 0 at the second impact 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values 𝑣 2 = (2.1) 2 +2(−9.8)(0) Work out terms 22.5𝑐𝑚 𝑣 2 =4.41 Square root 𝑣=2.1 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 𝑠=0 𝑎=−9.8 0.9m 𝑢=2.1 𝑡= ? 0.225m 𝑣= ? Approach speed: 2.1ms-1 Height of the second bounce Now we can calculate the rebound speed… Rebound speed: 1.05ms-1 𝑒= 𝑣 𝑢 Sub in values to find the rebound velocity 0.5= 𝑣 2.1 Multiply by 2.1 1.05=𝑣 22.5𝑐𝑚 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 𝑠= ? 𝑎=−9.8 0.9m 𝑢=1.05 𝑡= ? 0.225m 𝑣=0 ? Approach speed: 2.1ms-1 Height of the second bounce Like the first time, set v = 0 at the height of the bounce and use the newly calculated rebound speed Rebound speed: 1.05ms-1 𝑣 2 = 𝑢 2 +2𝑎𝑠 Sub in values (0) 2 = (1.05) 2 +2(−9.8)𝑠 Work out terms 0=1.1025−19.6𝑠 22.5𝑐𝑚 Add 19.6s 19.6𝑠=1.1025 Divide by 19.6 5.625𝑐𝑚 𝑠= 𝑚 4D

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 90+
𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 0.9m 0.225m 0.5625m Calculating the total distance travelled The ball falls 90cm It then bounces up to 22.5cm, and falls down 22.5cm again It then bounces up to 5.625cm, and falls down 5.625cm again This makes a sequence… 22.5𝑐𝑚 90+ … 5.625𝑐𝑚 90+2( …) 4D

𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Applet for collision demonstrations You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately. You can also solve problems relating to successive bounces on a horizontal plane. A tennis ball, which may be modelled as a particle, is dropped from rest at a height of 90cm onto a smooth horizontal plane. The coefficient of restitution between the ball and the plane is 0.5. Assume there is no air resistance and the ball falls freely under gravity at a right angle to the plane. Find the height to which the ball rebounds after the first bounce Find the height to which the ball bounces after the second bounce Find the total distance travelled by the ball before it comes to rest 0.9m 0.225m 0.5625m 90+2( …) The part in the bracket forms a geometric sequence  You will need to use the formula for the sum to infinity of a geometric sequence (from C2) 𝑎=𝐹𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝑆 ∞ = 𝑎 1−𝑟 𝑟=𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜(𝑑𝑖𝑣𝑖𝑑𝑒 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡𝑒𝑟𝑚 𝑏𝑦 𝑓𝑖𝑟𝑠𝑡)  The first term is 22.5  The common ratio is 0.25 22.5𝑐𝑚 −0.25 = 150 5.625𝑐𝑚 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒 150𝑐𝑚 4D

Teachings for Exercise 4E

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 3=𝑦−𝑥 4E
Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact 3=𝑦−𝑥 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two spheres have equal radii and masses 3kg and 5kg respectively. A and B move towards each other along the same straight line on a smooth horizontal surface with velocities 3ms-1 and 2ms-1 respectively. If the coefficient of restitution is 3/5, find the velocities of the spheres after the collision Find the loss of kinetic energy due to the impact 3 2 x y A B A B 3kg 5kg 3kg 5kg Approach = 5 Separation y - x 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 Sub in values 3 5 = 𝑦−𝑥 5 Multiply by 5 3=𝑦−𝑥 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact 3=𝑦−𝑥 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two spheres have equal radii and masses 3kg and 5kg respectively. A and B move towards each other along the same straight line on a smooth horizontal surface with velocities 3ms-1 and 2ms-1 respectively. If the coefficient of restitution is 3/5, find the velocities of the spheres after the collision Find the loss of kinetic energy due to the impact 3 2 x y −1=3𝑥+5𝑦 A B A B 3kg 5kg 3kg 5kg 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values 3 3 − 5 2 = 3 𝑥 +(5)(𝑦) Calculate −1=3𝑥+5𝑦 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact 3=𝑦−𝑥 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two spheres have equal radii and masses 3kg and 5kg respectively. A and B move towards each other along the same straight line on a smooth horizontal surface with velocities 3ms-1 and 2ms-1 respectively. If the coefficient of restitution is 3/5, find the velocities of the spheres after the collision Find the loss of kinetic energy due to the impact 3 2 x y −1=3𝑥+5𝑦 A B A B 𝑥=−2 𝑦=1 3kg 5kg 3kg 5kg Rearrange 1) 3=𝑦−𝑥 3+𝑥=𝑦 2) −1=3𝑥+5𝑦 −1=3𝑥+5𝑦 Replace y with the equivalent expression −1=3𝑥+5( ) 3+𝑥 Multiply out bracket −1=3𝑥+15+5𝑥 𝑥=−2𝑚 𝑠 −1 𝑦=1𝑚 𝑠 −1 Rearrange/Simplify −16=8𝑥 Divide by 8 −2=𝑥 Use this to find the value of y 1=𝑦 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact 3=𝑦−𝑥 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two spheres have equal radii and masses 3kg and 5kg respectively. A and B move towards each other along the same straight line on a smooth horizontal surface with velocities 3ms-1 and 2ms-1 respectively. If the coefficient of restitution is 3/5, find the velocities of the spheres after the collision Find the loss of kinetic energy due to the impact 3 2 2 x 1 y −1=3𝑥+5𝑦 A B A B 𝑥=−2 𝑦=1 3kg 5kg 3kg 5kg To calculate the kinetic energy lost, calculate the kinetic energy before and after the impact The direction of motion does not matter Kinetic energy before For A For B 𝐾𝐸= 1 2 𝑚 𝑣 2 𝐾𝐸= 1 2 𝑚 𝑣 2 Sub in values Sub in values 𝑥=−2𝑚 𝑠 −1 𝑦=1𝑚 𝑠 −1 𝐾𝐸= 1 2 (3)( 3) 2 𝐾𝐸= 1 2 (5)( 2) 2 Calculate Calculate 𝐾𝐸=13.5𝐽 𝐾𝐸=10𝐽 Kinetic energy before = 23.5J 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact 3=𝑦−𝑥 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two spheres have equal radii and masses 3kg and 5kg respectively. A and B move towards each other along the same straight line on a smooth horizontal surface with velocities 3ms-1 and 2ms-1 respectively. If the coefficient of restitution is 3/5, find the velocities of the spheres after the collision Find the loss of kinetic energy due to the impact 3 2 2 1 −1=3𝑥+5𝑦 A B A B 𝑥=−2 𝑦=1 3kg 5kg 3kg 5kg To calculate the kinetic energy lost, calculate the kinetic energy before and after the impact The direction of motion does not matter Kinetic energy after For A For B 𝐾𝐸= 1 2 𝑚 𝑣 2 𝐾𝐸= 1 2 𝑚 𝑣 2 Sub in values Sub in values 𝑥=−2𝑚 𝑠 −1 𝑦=1𝑚 𝑠 −1 𝐾𝐸= 1 2 (3)( 2) 2 𝐾𝐸= 1 2 (5)( 1) 2 Calculate Calculate 𝐾𝐸=6𝐽 𝐾𝐸=2.5𝐽 Kinetic energy before = 23.5J Kinetic energy after = 8.5J Kinetic energy lost = 15J 4E

Collisions 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 4E
Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse A gun of mass 600kg fires a shell of mass 12kg horizontally, with velocity 20ms-1. Find the velocity of the gun after the shell has been fired Find the total kinetic energy generated on firing Show that the ratio of the energy of the gun to the energy of the shell is equal to the ratio of the speed of the gun to the speed of the shell x 20 G S G S 600kg 12kg 600kg 12kg 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values (12)(0)=(600)(𝑥)+(12)(20) Calculate terms 0=600𝑥+240 𝑥=−0.4𝑚 𝑠 −1 Subtract 240 −240=600𝑥 Divide by 600 −0.4=𝑥 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse A gun of mass 600kg fires a shell of mass 12kg horizontally, with velocity 20ms-1. Find the velocity of the gun after the shell has been fired Find the total kinetic energy generated on firing Show that the ratio of the energy of the gun to the energy of the shell is equal to the ratio of the speed of the gun to the speed of the shell 0.4 x 20 G S G S 600kg 12kg 600kg 12kg  There was no kinetic energy to begin with as both objects were stationary  Calculate the Kinetic energy of both objects after the shell has been fired Kinetic energy after firing 𝑥=−0.4𝑚 𝑠 −1 For the gun For the shell 𝐾𝐸= 1 2 𝑚 𝑣 2 𝐾𝐸= 1 2 𝑚 𝑣 2 𝐾𝐸(𝑔𝑢𝑛)=48𝐽 𝐾𝐸(𝑠ℎ𝑒𝑙𝑙)=2400𝐽 Sub in values Sub in values 𝐾𝐸= 1 2 (600)( 0.4) 2 𝐾𝐸= 1 2 (12)( 20) 2 Calculate Calculate 𝐾𝐸=48𝐽 𝐾𝐸=2400𝐽 The total kinetic energy will be 2448J  This has been generated from the chemical energy in the firing of the gun 4E

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before impact After impact You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse A gun of mass 600kg fires a shell of mass 12kg horizontally, with velocity 20ms-1. Find the velocity of the gun after the shell has been fired Find the total kinetic energy generated on firing Show that the ratio of the energy of the gun to the energy of the shell is equal to the ratio of the speed of the gun to the speed of the shell 0.4 20 G S G S 600kg 12kg 600kg 12kg Ratio of energy of gun to energy of shell 48:2400 Divide by 48 1:50 𝑥=−0.4𝑚 𝑠 −1 Ratio of velocity of gun to velocity of shell 𝐾𝐸(𝑔𝑢𝑛)=48𝐽 𝐾𝐸(𝑠ℎ𝑒𝑙𝑙)=2400𝐽 0.4:20 Divide by 0.4 1:50 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse A tennis ball of mass 0.2kg is moving with velocity (-12i – 2j)ms-1 when it is struck by a tennis racquet. This is modelled as an impulse of (5.6i + 2.8j)Ns. Find the resulting velocity of the tennis ball and the kinetic energy gained by the ball as a result of the impact. 𝑰=𝑚𝒗−𝑚𝒖 Sub in values (5.6𝒊+2.8𝒋)=(0.2)(𝒗)−(0.2)(−12𝒊 −2𝒋) Multiply out the bracket 5.6𝒊+2.8𝒋=0.2𝒗+2.4𝒊+0.4𝒋 Subtract 2.4i and 0.4j 3.2𝒊+2.4𝒋=0.2𝒗 Multiply by 5 16𝒊+12𝒋=𝒗 Initial velocity: (-12i – 2j)ms-1 Final velocity: (16i + 12j)ms-1 4E

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse A tennis ball of mass 0.2kg is moving with velocity (-12i – 2j)ms-1 when it is struck by a tennis racquet. This is modelled as an impulse of (5.6i + 2.8j)Ns. Find the resulting velocity of the tennis ball and the kinetic energy gained by the ball as a result of the impact. We need to know the magnitude of the velocity before and after  Use Pythagoras’ Theorem Before: 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = (−12) 2 + (−2) 2 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 148 After: Initial velocity: (-12i – 2j)ms-1 𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = (16) 2 + (12) 2 = √148 ms-1 𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =20 Final velocity: (16i + 12j)ms-1 = 20 ms-1 4E

𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse A tennis ball of mass 0.2kg is moving with velocity (-12i – 2j)ms-1 when it is struck by a tennis racquet. This is modelled as an impulse of (5.6i + 2.8j)Ns. Find the resulting velocity of the tennis ball and the kinetic energy gained by the ball as a result of the impact. Now we can calculate the Kinetic energy before and after the impact: Before: 𝐾𝐸= 1 2 𝑚 𝑣 2 Sub in values 𝐾𝐸= 1 2 (0.2)( √148) 2 Calculate 𝐾𝐸=14.8𝐽 Initial velocity: (-12i – 2j)ms-1 After: = √148 ms-1 𝐾𝐸= 1 2 𝑚 𝑣 2 Sub in values Final velocity: (16i + 12j)ms-1 𝐾𝐸= 1 2 (0.2)( 20) 2 = 20 ms-1 Calculate 𝐾𝐸=40𝐽 𝑇ℎ𝑒 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙 𝑖𝑠 25.2𝐽 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before the jerk After the jerk You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two particles, A and B, of mass 200g and 300g respectively, are connected by a light inextensible string. The particles are side-by-side on a smooth floor and A is projected with speed 6ms-1 away from B. When the string become taut, particle B is jerked into motion and A and B then move a common speed in the direction of A’s original motion. Find: The common speed of the particles after the string becomes taut The loss of kinetic energy as a result of the jerk 6 v v B A B A 300g 200g 300g 200g 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Sub in values = 0.3 𝑣 +(0.2)(𝑣) Simplify 1.2=0.5𝑣 Multiply by 2 2.4=𝑣 𝑣=2.4𝑚 𝑠 −1 4E

Applet for collision demonstrations 𝑒= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑚 1 𝒖 1 + 𝑚 2 𝒖 2 = 𝑚 1 𝒗 1 + 𝑚 2 𝒗 2 Collisions 𝑒= 𝑣 𝑢 𝑰=𝑚𝒗−𝑚𝒖 𝑰=𝑭𝑡 Before the jerk After the jerk You can solve problems which ask you to find the change in energy due to an impact of the application of an impulse Two particles, A and B, of mass 200g and 300g respectively, are connected by a light inextensible string. The particles are side-by-side on a smooth floor and A is projected with speed 6ms-1 away from B. When the string become taut, particle B is jerked into motion and A and B then move a common speed in the direction of A’s original motion. Find: The common speed of the particles after the string becomes taut The loss of kinetic energy as a result of the jerk 6 2.4 v 2.4 v B A B A 300g 200g 300g 200g KE before the jerk 𝐾𝐸= 1 2 𝑚 𝑣 2 Sub in values – particle A is the only one with a velocity 𝐾𝐸= 1 2 (0.2)( 6) 2 Calculate 𝐾𝐸=3.6𝐽 KE after the jerk 𝐾𝐸= 1 2 𝑚 𝑣 2 Sub in values – model the two joined particles as a single one 𝐾𝐸= 1 2 (0.5)( 2.4) 2 𝑣=2.4𝑚 𝑠 −1 Calculate 𝐾𝐸=1.44𝐽 𝑇ℎ𝑒 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑗𝑒𝑟𝑘 𝑖𝑠 2.16𝐽 4E

Summary We have seen how to model collisions of particles including the use of vectors We have learnt how to use Newton’s Law of Restitution We have seen problems involving the sum of an infinite Geometric series We have also solved problems involving the change in Kinetic energy

Collisions.

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Collisions.

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Presentation on theme: "Collisions."— Presentation transcript:

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