2. If you apply a force to an object, the effect it has depends on the mass of the object and for how long you exert the force. You apply a constant force, F Newtons for a time t seconds in the direction of motion to a particle of mass m kg. The particle accelerates at a ms -2. The velocity of the particle increases from u ms -1 to v ms -1.

3. You have v = u + at From Newton’s second law you know that F = ma Ft = mv – mu

4. Because force is a vector quantity, impulse is also a vector quantity The IMPULSE of a force F applied for a time t is defined as Ft Impulse = force x time = Ft The SI unit of impulse is the newton second (Ns) For a particle of mass m moving with velocity v, Momentum = mv The SI unit of momentum is the newton second (Ns) In general Impulse = change in momentum Ft = mv – mu

5. Example 1 A spacecraft of mass 120kg is travelling in a straight line at a speed of 4ms -1. Its rocket is fired for 5 seconds, exerting a force of 150N. Find the new velocity of the spacecraft if the force was directed: (a)in the direction of motion (b)in the direction opposite to motion u = 4ms -1 150N (a) Impulse = Change in Momentum Ft = mv – mu 150 x 5 = 120v – 120 x 4 v = 10.25ms -1 u = 4ms -1 150N (b) Impulse = Change in Momentum Ft = mv – mu -150 x 5 = 120v – 120 x 4 v = -2.25ms -1

6. Example 2 A batsman hits a ball of mass 0.15kg at a speed of 40ms -1. The ball moves in a straight line with constant speed until it strikes a net at right angles and is brought to rest in 0.5 seconds. Find the magnitude of the impulse and the average force exerted by the netting on the ball. Mass = 0.15kg 40ms -1 0ms -1 F Momentum = mass x velocity Momentum = 0.15 x 40 Momentum = 6Ns Momentum = mass x velocity Momentum = 0.15 x 0 Momentum = 0Ns Impulse = Change in Momentum = 6Ns Impulse = Force x time 6 = F x 0.5 F = 12N

7. Example 3 A particle of mass 2kg, travelling with a velocity of (3i + 5j)ms -1, is given an impulse of (2i – 4j)Ns. Find its new velocity. Impulse = Change in Momentum Impulse = mv – mu 2i – 4j = 2v – 2(3i + 5j) 2i – 4j = 2v – 6i – 10j 8i + 6j = 2v v = (4i + 3j)ms -1

8. Example 4 Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms -1 towards Monica. She kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms -1 and with an elevation of 40° to the horizontal. Find the magnitude and direction of the impulse of Monica’s kick. Mass = 0.8kg 5ms -1 Mass = 0.8kg 8ms -1 40° 5i ms -1 (-8cos40i + 8sin40j) ms -1 Impulse = Change in Momentum Impulse = mv – mu Impulse = 0.8(-8cos40i + 8sin40j) – (0.8 x 5i) Impulse = -4.9027i + 4.1138j – 4i Impulse = -8.9027i + 4.1138j

9. Example 4 Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms -1 towards Monica. She kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms -1 and with an elevation of 40° to the horizontal. Find the magnitude and direction of the impulse of Monica’s kick. Impulse = -8.9027i + 4.1138j 8.9027 4.1138 θ I

10. Conservation of linear momentum We will now study the motion of systems comprising of two or more bodies, which are free to move separately A system consists of two bodies, A and B, which have momentum M A and M B respectively The total momentum of the system = (M A + M B ) Ns Body A now exerts a force, F N on body B for a time t seconds. Body B receives an impulse J = Ft Ns By Newton’s third law, B exerts a force, -F N on A, also for time t secs Body A receives an impulse –J Ns.

11. We know that Impulse = change in momentum The new momentum of A = (M A – J) Ns The new momentum of B = (M B + J) Ns The total momentum of the system = (M A – J) + (M B + J) = M A + M B i.e. the total momentum has not been changed The forces described here are internal forces to the system. The total momentum of a system can only be changed by the application of an external force. This is the principle of conservation of linear momentum.

12. The principle of conservation of linear momentum The total momentum of a system in a particular direction remains constant unless an external force is applied in that direction In the system below: Total momentum before collision = Total momentum after collision m A u A + m B u B = m A v A + m B v B

13. Example 5 A particle of mass 4kg, travelling at a speed of 6ms -1, collides with a second particle, of mass 3kg and travelling in the opposite direction at a speed of 2ms -1. After the collision, the first particle continues in the same direction but with its speed reduced to 1ms -1. Find the velocity of the second particle after the collision. 6ms -1 A B 2ms -1 4kg 3kg 1ms -1 A B v Before After m A u A + m B u B = m A v A + m B v B (4 x 6) + (3 x -2) = (4 x 1) + 3v 18 = 4 + 3v v = 14 / 3 ms -1

14. Example 6 A body of mass 5kg, travelling with a speed of 6ms -1, collides with a body B, of mass 3kg, travelling in the same direction with a speed of 4ms -1. On colliding, the two bodies coalesce. Find the velocity of the combined body after the collision. 6ms -1 A B 4ms -1 5kg 3kg Before v After 8kg A+B m A u A + m B u B = m A+B v (5 x 6) + (3 x 4) = 8v 42 = 8v v = 5.25ms -1

15. Example 7 A railway truck of mass 4 tonnes, travelling along a straight horizontal trail at a speed of 4ms -1, meets another truck, of mass 2 tonnes, travelling in the opposite direction at a speed of 5ms -1. They collide and become coupled together. Find their velocity after the collision. 4ms -1 A B 5ms -1 4000kg 2000kg Before v After 6000kg A+B m A u A + m B u B = m A+B v (4000 x 4) + (2000 x -5) = 6000v 6000 = 6000v v = 1ms -1 (in direction of A originally)

16. Example 8 A body of mass 4kg travelling with a velocity of (3i + 2j)ms -1 collides and coalesces with a second body of mass 3kg travelling with velocity (i – 3j)ms -1. Find their common velocity after impact. m A u A + m B u B = m A+B v 4(3i + 2j) + 3(i – 3j) = 7v 15i – j = 7v

17. Example 9 Two particles, A and B, of mass 3kg and 2kg respectively, lie at rest on a smooth horizontal table. They are connected by a light inextensible string which is initially slack. B starts to move at a speed of 8ms -1 in the direction AB. Find the common velocity of the particles immediately after the string becomes taut, and the impulse received by each of the particles. A B 8ms -1 3kg 2kg Before A B v 3kg 2kg After Momentum Before = 2 x 8= 16Ns Momentum After = 5v Momentum Before Momentum After = 16 = 5v v = 3.2ms -1 Impulse = Change in momentum (momentum after – momentum before) (consider just one of the particles) A = (3 x 3.2) – (3 x 0) = 9.6Ns B = (2 x 3.2) – (2 x 8) = -9.6Ns Impulse = 9.6Ns

18. Example 10 A bullet of mass 50 grams is fired horizontally from a gun of mass 1kg, which is free to move. The bullet is fired with a velocity of 250ms -1. Find the speed with which the gun recoils. m = 1kg m = 0.05kg 0ms -1 250ms -1 v m = 1kg m = 0.05kg BEFORE AFTER Momentum Before = 0Ns Momentum After = (250 x 0.05) + (1 x –v) = 12.5 – v Momentum Before = Momentum After 0 = 12.5 – v v = 12.5ms -1

19. Example 11 A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms -1. The gun rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary before the shot is fired. Find the distance the gun will move as a result of firing the shell. BEFORE AFTER 804g800g 4g u = 0ms -1 400ms -1 v R 0.6R Momentum Before = 0Ns Momentum After = (400 x 4) + (800 x –v) = 1600 – 800v Momentum Before = Momentum After 0 = 1600 – 800v v = 2ms -1

20. Example 11 A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms -1. The gun rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary before the shot is fired. Find the distance the gun will move as a result of firing the shell. 800g u = 2ms -1 R 0.6R v = 0ms -1 s a = ? F = ma -0.6R = 800a Resolving vertically R = 800g -0.6(800g) = 800a a = -5.88ms -2 u = 2ms -1 v = 0ms -1 a = -5.88ms -2 s = ? v 2 = u 2 + 2as 0 2 = 2 2 + (2 x -5.88 x s) s = 0.34m (34cm)