Question of Physics

Question

A ball of mass m collides with a wall with initial velocity -v₀ j

(and then it rebounds). The force acting on the ball by the wall during the collision is given by F = (At + Bt²) j where A and B are positive constants. Find the velocity of the ball after the collision. The ball is in contact with the wall 10^-2 seconds (Assume that the gravitational force is negligible compared with force F).

Answer 1

We use the equality between the impulse and change in momentum.

The impulse, being the integral of F from time 0 to 10^-2, is

A(10^-2)²/2 + B(10^-2)³/3 = A10^-4/2 + B10^-6/3.

Let the velocity of the ball after the collision be v.

The momentum before collision is -mv₀, and after collision mv.

Setting the difference in momentum equal to the impulse:

mv + mv₀ = A10^-4/2 + B10^-6/3

v = (A10^-4/2 + B10^-6/3)/m - v0