Question

A ball travels along a straight path, perpendicular to a wall. The ball bounces off the wall and returns with the same magnitude of momentum, p, with respect to a ground observer at rest. What expression shows the change in momentum as observed by the ground observer?.

Answer 1

The change in momentum (Δp) for a ball bouncing off a wall and returning along its original path with the same speed is -2mv, which indicates a reversal of the momentum in the x-direction.

Step-by-step explanation:

The change in momentum of a ball bouncing off a wall, as observed by a ground observer, can be calculated by considering the differences in the ball's momentum before and after the collision with the wall. Since momentum is a vector quantity, which includes both magnitude and direction, the change in momentum (Δp) involves a change in the direction of the ball's velocity. In this scenario, the ball's speed (u) and mass (m) remain the same before and after the collision, but the direction of the velocity changes by 180 degrees when the ball bounces back. Consequently, the initial momentum (p) is mv in the +x direction and the final momentum (p') is mv in the -x direction. The expression for the change in momentum is Δp = p' - p, which simplifies to Δp = (-mv) - (mv) or Δp = -2mv.