Question

PLEASE HELPP!!!

A 0.75kg ball is thrown against a wall at 30m/s. The ball is in
contact with the wall for only 0.05sec and bounces back at 10m/s.
Calculate initial momentum of the ball and momentum imparted
by the wall.

Answer 1

a) The initial momentum of the ball is 22.5 kg·m/s

b) The magnitude of the momentum imparted by the ball is 30 kg·m/s

Step-by-step explanation:

The question is based on change in momentum

The mass of the ball, m = 0.75 kg

The velocity with which the ball is thrown against the wall, u = 30 m/s

The time duration it takes while the ball is in contact with the wall, Δt = 0.05 sec

The velocity of the ball as it bounce back, v = -10 m/s (The ball moves in the opposite direction)

a) The initial momentum of the ball,
`P_(Initial)` = m × u = 0.75 kg × 30 m/s = 22.5 kg·m/s

b) The final momentum of the ball,
`P_(Final)` = m × v = 0.75 kg × (-10 m/s) = -7.5 kg·m/s

The momentum imparted by the ball, ΔP = The final momentum - The initial momentum

∴ ΔP =
`P_(Final)` -
`P_(Initial)` = -7.5 kg·m/s - 22.5 kg·m/s = -30 kg·m/s

The magnitude of the momentum imparted by the ball,
`\left | \Delta P \right |` = 30 kg·m/s