1. Momentum

2. Impulse

3. Impulse
A collision is a short-duration interaction between two objects.
Collisions may appear instantaneous, but their duration, however small, is significant.

4. Impulse
Applied force increases and then decreases in magnitude throughout the collision’s duration. (Think kicking a soccer ball).
A large force like this exerted during a short interval of time is called an impulsive force.

5. Impulse
An impulsive force of greater magnitude (taller force curve) or a force applied over a longer duration (wider force curve) has a greater effect.
A taller or wider force curve also has a larger area under the curve.
This is area is called the impulse, J.

6. Impulse
Impulsive forces can be complex, and the shape of F-t graphs can often change in complicated ways.
Because of this, it is often useful to think of the collision in terms of an average force, Favg.
𝑖𝑚𝑝𝑢𝑙𝑠𝑒 𝐽= 𝐹 𝑎𝑣𝑔 ∆𝑡

7. Momentum and the Impulse-Momentum Theorem

8. Momentum
Effect of an impulsive force also depends on the mass of the object.
e.g., kicking a heavy object will change its velocity much less than giving the same kick to a light object.
Consider an object moving with initial velocity, vo.
You kick this object and deliver an impulse J = Favg∆t
After impulse, the object now moves with a final velocity, v.
How is this final velocity related to the initial velocity?

9. Momentum
From Newton’s second law, the average acceleration of said object during the time interval of the kick is:
𝑎 𝑎𝑣𝑔 = 𝐹 𝑎𝑣𝑔 𝑚
Average acceleration is related to the change in velocity by:
𝑎 𝑎𝑣𝑔 = ∆𝑣 ∆𝑡 = 𝑣− 𝑣 𝑜 ∆𝑡
Combining the two equations for average acceleration:
𝐹 𝑎𝑣𝑔 𝑚 = 𝑣− 𝑣 𝑜 ∆𝑡
Or:
𝐹 𝑎𝑣𝑔 ∆𝑡=𝑚𝑣− 𝑚𝑣 𝑜

10. Momentum 𝐹 𝑎𝑣𝑔 ∆𝑡=𝑚𝑣− 𝑚𝑣 𝑜
The left side of the equation is impulse, J. The right side is the change in the quantity mv (mass cannot change). This quantity is called the momentum, p of the object.
𝑝=𝑚𝑣
Momentum is a vector quantity that points in the same direction as the velocity vector, so it too has components, x- and y-.

11. The Impulse-Momentum Theorem
𝐹 𝑎𝑣𝑔 ∆𝑡=𝑚𝑣− 𝑚𝑣 𝑜
This expression relating impulse and momentum can be written in corresponding terms as:
𝐽=𝑝− 𝑝 𝑜 =∆𝑝
This expression is known as the impulse-momentum theorem. It states that an impulse delivered to an object causes the object’s momentum to change.

12. EXAMPLE 1 Calculating the change in momentum
A ball of mass m = 0.25 kg rolling to the right at 1.3 m/s strikes a wall and rebounds to the left at 1.1 m/s. What is the change in the ball’s momentum? What is the impulse delivered to it by the wall?

13. EXAMPLE 1 Calculating the change in momentum
Initial momentum is:
𝑝 𝑜 =𝑚 𝑣 𝑜 = 0.25 𝑘𝑔 1.3 𝑚/𝑠 =0.325 𝑘𝑔∙𝑚/𝑠
After the ball rebounds, the momentum is:
𝑝=𝑚 𝑣 𝑜 = 0.25 𝑘𝑔 −1.1 𝑚/𝑠 =−0.28 𝑘𝑔∙𝑚/𝑠
So change in momentum is:
∆𝑝=𝑝− 𝑝 𝑜 = −0.28 − 0.33 =−0.61 𝑘𝑔∙𝑚/𝑠
By the impulse-momentum theorem, the impulse delivered to the ball by the wall is equal to this change, so:
𝐽=∆𝑝=−0.61 𝑘𝑔∙𝑚/𝑠

14. EXAMPLE 2 A well-hit ball
A baseball of mass 0.14 kg has an initial velocity of 38 m/s as it approaches a bat. The bat applies a force that is much larger that the weight of the ball, and the ball departs from the bat with a final velocity of 58 m/s. (a) Determine the impulse applied to the ball by the bat. (b) Assuming that the time of contact is 1.6 x 10-3 s, find the average net force exerted on the ball by the bat.

15. EXAMPLE 2 A well-hit ball
(a) Impulse-momentum theorem says impulse is equal to:
𝐽=𝑚𝑣−𝑚 𝑣 𝑜
= − −38
=13.4 𝑘𝑔∙𝑚/𝑠
(b) Now that the impulse is known, the contact time can be used to find the average force:
𝐽= 𝐹 𝑎𝑣𝑔 ∆𝑡
𝐽 ∆𝑡 = 𝐹 𝑎𝑣𝑔
× 10 −3 =8400 𝑁

16. EXAMPLE 3 A Rain Storm
During a storm, rain comes straight down with a velocity of - 15 m/s and hits the roof of a car perpendicularly. The mass of rain per second that strikes the car roof is kg/s. Assuming that the rain comes to rest upon striking the car, find the average force exerted by the rain on the roof.

17. EXAMPLE 3 A Rain Storm
Favg needed to reduce the rain’s velocity from vo = -15 m/s to v = 0 m/s is given by:
𝐹= 𝑚𝑣−𝑚𝑣 𝑜 ∆𝑡 =− 𝑚 ∆𝑡 𝑣 𝑜
The term 𝑚 ∆𝑡 is the mass of rain per second that strikes the roof, so that 𝑚 ∆𝑡 =0.060 kg/s. Thus, the average force acting on the rain is
𝐹=− −15 =0.90 N

18. Stopping Objects
Think of a fast-moving object that comes to an abrupt stop by collision. In this example, the object has momentum po just before impact and zero momentum after impact (p = 0).
The impulse-momentum theorem states that:
𝐽= 𝐹 𝑎𝑣𝑔 ∆𝑡=∆𝑝=𝑝− 𝑝 𝑜 =− 𝑝 𝑜
Or:
𝐹 𝑎𝑣𝑔 =− 𝑝 𝑜 ∆𝑡
This expression is to say that the average force needed to stop an object is inversely proportional to the duration of the collision.

19. Stopping Objects
If the duration of the collision can be increased, the force of the impact will be decreased.
This is the principal used in most impact-lessening techniques, like the water barrels on the freeway, or bending your knees when coming down from free-fall.

20. Total Momentum
If we have more than one object moving—a system of particles—then the system as a whole has an overall momentum.
The total momentum P (note the capital P) of a system of particles is the vector sum of the momenta of the individual particles:
𝑃= 𝑝 1 + 𝑝 2 + 𝑝 3 …

21. Conservation of Momentum

22. Newton’s Laws and Momentum
Impulse-momentum theorem was derived from Newton’s second law. It serves as an alternative way of looking at second law, but in the context of only one particle at a time.
Consider now two objects instead. When two objects collide, they exert forces on each other in often complicated ways, so using only NSL to predict the behavior of these collisions would be difficult.
Newton’s third law provides a simpler way of predicting the outcome of a collision.

23. Action-Reaction and Momentum
Picture two balls headed toward each other, colliding and then bouncing apart.
The forces during the collision, when the balls are interacting, are the action-reaction pair F1 on 2 and F2 on 1.

24. Action-Reaction and Momentum
During the collision, the impulse J2 delivered to ball 2 by ball 1 is the average value of F1 on 2 multiplied by the collision time Δt.
Likewise, the impulse J1 delivered to ball 1 by ball 2 is the average value of F2 on 1 multiplied by Δt.
F1 on 2 and F2 on 1 are an action-reaction pair equal in magnitude but opposite in direction, so J1 = -J2.

25. Action-Reaction and Momentum
Impulse-momentum theorem says Δp is equal to J, so Δp for either ball is also equal in magnitude but opposite in sign.
In other words, if ball 1’s momentum increases by a certain amount during the collision, ball 2’s momentum will decrease by exactly the same amount.
This implies that total momentum P = p1 + p2 of the two balls is unchanged by the collision.
Because it doesn’t change during the collision, we say momentum is conserved.

26. Law of Conservation of Momentum
The same conservation of momentum holds true for systems containing any number of objects.
Forces that act only between particles within the system are called internal forces.

27. Law of Conservation of Momentum
Most systems are also subject to forces from sources outside the system. These forces are called external forces, and they can change the momentum of the system.
Say there are three particles in a system, and they are all being acted on by an external force. Each would have their momentum changed by a value ∆𝑝 so that the change in the total momentum of the system is
∆𝑃=∆ 𝑝 1 +∆ 𝑝 2 +∆ 𝑝 3

28. Law of Conservation of Momentum
By impulse-momentum theorem, change in momentum is
∆𝑝= 𝐹 𝑎𝑣𝑔 ∆𝑡
So total momentum of the system is also changing by
∆𝑃= 𝐹 𝑒𝑥𝑡 𝑜𝑛 1 ∆𝑡+ 𝐹 𝑒𝑥𝑡 𝑜𝑛 2 ∆𝑡+ 𝐹 𝑒𝑥𝑡 𝑜𝑛 3 ∆𝑡
=( 𝐹 𝑒𝑥𝑡 𝑜𝑛 1 + 𝐹 𝑒𝑥𝑡 𝑜𝑛 2 + 𝐹 𝑒𝑥𝑡 𝑜𝑛 3 )∆𝑡
= 𝐹 𝑛𝑒𝑡 ∆𝑡
where 𝐹 𝑛𝑒𝑡 is the net force due to external forces.
So, if the net external forces acting on a system is equal to zero, the total momentum of the system does not change.
Law of Conservation of Momentum - The total momentum P of an isolated system is a constant. Interaction within the system does not change the system's total momentum. It is conserved.
𝑃= 𝑃 𝑜

29. EXAMPLE 4 Speed of ice skaters pushing off
Two ice skaters, Sandra and David, stand facing each other on frictionless ice. Sandra has a mass of 45 kg, David a mass of 80 kg. They then push off from each other. After the push, Sandra moves off at a speed of 2.2 m/s. What is David’s speed?

30. EXAMPLE 4 Speed of ice skaters pushing off
We write Sandra’s initial momentum as 𝑝 𝑆,𝑜 = 𝑚 𝑆 𝑣 𝑆,𝑜
David’s initial momentum

31. Explosions
An explosion, where the particles of the system move apart after a brief, intense interaction, is the opposite of a collision.
Explosive forces (expanding spring, expanding hot gases) are internal forces, so if the system is isolated, its total momentum during the explosion will be conserved.

32. EXAMPLE 6 Recoil speed of a rifle
A 30 g ball is fired from a 1.2 kg spring-loaded toy rifle with a speed of 15 m/s. What is the recoil speed of the rifle?

33. EXAMPLE 6 Recoil speed of a rifle
𝑃= 𝑝 𝑏 + 𝑝 𝑟
Everything is at rest before the trigger is pulled, so initial momentum is zero.
After the trigger is pulled, the internal force of the spring pushes the ball down the barrel and pushes the rifle backward. Conservation of momentum gives
𝑃= 𝑚 𝑏 𝑣 𝑏 + 𝑚 𝑟 𝑣 𝑟 = 𝑃 𝑜 =0
Solving for the rifle’s velocity gives
𝑣 𝑟 =− 𝑚 𝑏 𝑚 𝑟 𝑣 𝑏 =− ×15=−0.38 𝑚/𝑠

34. Inelastic Collisions

35. Inelastic Collisions
Sometimes colliding objects bounce off each other. This type of collision is known as an elastic collision.
Colliding objects may also stick to each other. This is known as an inelastic collision.

36. EXAMPLE 7 Speeds in an inelastic glider collision
In a laboratory experiment, a 200g air-track glider and a 400 g air-track glider are pushed toward each other from opposite ends of the track. The gliders have Velcro tabs on their fronts so that they will stick together when they collide. The 200 g glider is pushed with an initial speed of 3.0 m/s. The collision causes it to reverse direction at 0.50 m/s. What was the initial speed of the 400 g glider?

37. EXAMPLE 7 Speeds in an inelastic glider collision
The law of conservation of momentum is
𝑃 𝑜 =𝑃
𝑚 1 𝑣 1,𝑜 + 𝑚 2 𝑣 2,𝑜 = 𝑚 1 + 𝑚 2 𝑣
Because the two gliders stick together and move at a common velocity.
Solving for the initial velocity of the 400 g glider
𝑣 2,𝑜 = 𝑚 1 + 𝑚 2 𝑣− 𝑚 1 𝑣 1,𝑜 𝑚 2
= −0.50 −(0.20)(3.0) 0.40
=−2.3 𝑚/𝑠

38. Momentum and Collisions in Two Dimensions

39. EXAMPLE 8 Analyzing a peregrine falcon strike
Peregrine falcons often grab their prey from above while both falcon and prey are in flight. A falcon, flying at 18 m/s, swoops down at a 45° angle from behind a pigeon flying horizontally at 9.0 m/s. The falcon has a mass of 0.80 kg and the pigeon a mass of 0.36 kg. What are the speed and direction of the falcon (now holding the pigeon) immediately after impact?

40. EXAMPLE 8 Analyzing a peregrine falcon strike
Start by finding the x- and y- components of the momentum of the system (falcon and pigeon) before the collision.
For the x-components we have
𝑃 𝑜,𝑥 = 𝑚 𝑓 𝑣 𝑓𝑥,𝑜 + 𝑚 𝑝 𝑣 𝑝𝑥,𝑜 = 𝑚 𝑓 𝑣 𝑓,𝑜 cos 𝜃 + 𝑚 𝑝 𝑣 𝑝
=(0.80)(18 cos 45°) +(0.36)(9.0)
=13.4 𝑘𝑔∙𝑚/𝑠
For the y-component we have
𝑃 𝑜,𝑦 = 𝑚 𝑓 𝑣 𝑓𝑦,𝑜 + 𝑚 𝑝 𝑣 𝑝𝑦,𝑜 = −𝑚 𝑓 𝑣 𝑓,𝑜 sin 𝜃 +0
= −18.0 sin 45°
=−10.2 𝑘𝑔∙𝑚/𝑠

41. EXAMPLE 8 Analyzing a peregrine falcon strike
After the collision, the two birds move with a common velocity that is directed at an angle.
X-component of the final momentum is