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Linear Momentum For an individual mass we define the linear momentum to be: From the 2 nd law we have: This is actually how Newton originally formulated the 2 nd law. The “ma” is a special case when m is not changing.

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Linear Momentum of a System of Particles For a system of many particles, we can define the “total linear momentum”: Then we can write: The total linear momentum is a result of the total mass moving with the velocity of the CM

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Let us differentiate the total momentum: If there are no external forces, then momentum is conserved: This is a vector equation. If there are not external forces in say the x-direction, then P xo = P xf even if there are external forces in the y-direction.

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Elastic Collisions in 1-D Elastic means KE stays the same:

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Dividing two equations gives (upon rearrangement): The relative velocity changes direction but keeps same magnitude. Only true in 1-D, elastic collisions. Example: Ping pong ball collides with stationary bowling ball. Multiply by M and subtract Bounces back with same speed

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Example: Bowling ball hits stationary ping pong ball. Multiply by M and add Barry Bonds uses a light bat

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A two dimensional collision

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Example: a 4 kg mass heading in the – y direction at 12 m/s collides and sticks to a 6 kg mass moving in the + x direction at 10 m/s. Find the magnitude and direction of the final velocity.

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Impulse For a constant force, let is define a vector quantity called impulse as the product of the force times the time over which it acts: In one dimension, we need only worry about the sign. If the force is not constant during the time over which the force acts, we define through an integral: Note the analogy to our definition of work. Of course a huge difference is work is a scalar and impulse is a vector. For any component, the impulse will be the area under the F i vs t graph.

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So why bother with impulse? Suppose we focus on the impulse delivered by the net force.

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Impulse – Momentum Theorem p = p f – p i = F dt tftf titi The impulse of the force F acting on a particle equals the change in the momentum of the particle. J = p

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Example: A 0.1 kg mass moving with a speed of 10 m/s along the x-axis collides head on with a stationary 0.1kg mass. The magnitude of the force between the two is shown below as a function of time. Find the final speed and direction of each mass. For 2 nd block: For 1 st block: Impulse is negative by Newton’s 3 rd Law

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