A 3 kg ball is thrown horizontally towards a wall with a speed of 5.0 m/s. The i

Question

A 3 kg ball is thrown horizontally towards a wall with a speed of 5.0 m/s. The initial velocity is chosen to be the positive x-direction for this question. The ball horizontally rebounds back from the wall with a speed of 3.0 m/s in the negative x- direction. The ball is in contact with the wall for 150 milliseconds. (a) What is the change in momentum, Delta p vector, of the ball? (b) What is the Impulse that the wall exerts on the ball, I vector_ball, wall? (c) Use the impulse-momentum theorem to find the force that the ball exerts on the wall, F vector_wall, ball? Bob and Linda are sweethearts that enjoy skating aim in arm. Bob has a mass of 85 kg and Linda has a mass of 60 kg. Initially Bob is standing still when Linda comes up behind him with a speed of 3.0 m/s. What is the couple's final speed?

Explanation / Answer

1.(a) Initial momnetum P1 = mv = 3kg(5m/s) = 15kg-m/s along +x direction

Final momentum is P2 = mv = 3kg(-3m/s) = -9kg-m/s, so momentum is 9kg-m/s along -x direction.

So change in momentum is p = P2 - P1 = (-9kg-m/si) - (15kg-m/si) = -24kg-m/si, where i is unit vector along +x direction.

So, p = -24kg-m/si, or 24kg-m/s in -x direction

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1.(b) Iball,wall = p

or Iball,wall = -24kg-m/si, or 24kg-m/s in -x direction

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1.(c) As per the impulse momentum theorem

Fball,wall = p/t =  Iball,wall/t

or   Fball,wall = (-24kg-m/si)/(150ms) = (-24kg-m/si)/(0.15s)

or  Fball,wall = -160 Ni, or 160N in -x direction,

So, Fwall,ball = -Fball,wall = 160Ni,  or 160N in +x direction,

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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