Lecture 22: Momentum and Impulse
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Lecture Objectives 1. Define momentum and impulse 2. Relate the momentum, impulse, force, and time of contact in a system.
Momentum is a quantity associated with the motion of a system.
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Example A 1000kg truck is moving 0.1m/s. What should be the speed of a 5kg skateboard so that they would have the same momentum? ptruck = mv = 100kg m/s;
v = (100kgm/s)/5kg = 20m/s
Change in momentum
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Impulse-‐Momentum Theorem
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Impulse-‐Momentum Theorem
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Question: If you were in a car that was out of control and had to choose between hitting a haystack or a concrete wall. (Assuming everything is okay in your life), what would you choose? A. Haystack
B. Concrete wall
If the change in momentum occurs over a short time, the force of impact is large.
Impulse is the change in momentum! J = Ft = mv2 – mv1
If the change in momentum occurs over a long time, the force of impact is small.
Impulse is the change in momentum! J = Ft = mv2 – mv1
Haystack Physics helps you to understand why hitting a soft object is entirely different from hitting a hard one.
Force of impact will be small if momentum change occurs in a long time: J = Ft
Note: Impulse is the change in momentum!
J = Ft = mv2 – mv1
Impulse and Average Force
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Momentum and Kinetic Energy
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Sample Problem
Suppose you throw a ball with a mass of 0.40kg against a brick wall. It hits the wall moving horizontally to the left at 30m/s and rebounds horizontally to the right at 20m/s. (a) Find the impulse of the net force on the ball during its collision with the wall. (b) If the ball is in contact with the wall in 0.01s, find the average horizontal force that the wall exerts on the ball during the impact.
Given: m = 0.40kg v1x = -‐30m/s v1x = 20m/s Δt = 0.01s Required: J and F
The initial and final x-‐components of momentum of the ball are:
The x-‐component of impulse is equal to the change in x-‐momentum:
To solve for the force during the impact, use the formula for impulse:
Given: m = 0.40kg v1x = -‐30m/s v1x = 20m/s Δt = 0.01s Required: J and F
Center of Mass
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A 1250 kg automobile is moving from east to west at a speed of 25.0 m/s and comes to a stop in 10.0 s.
1) Find the impulse on the automobile. 2) Find the magnitude of the net force acting on the automobile in the 10.0 s interval.
Seatwork
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Seatwork 4 & 5 (Young and Freedman, 8.8)A baseball has mass 0.145 kg. 4) If the velocity of the pitched ball has magnitude of 45.0 m/s and the batted ball’s velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball (J).5) If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by the bat. (1ms = 0.001s)