Question

A racquet ball with mass m = 0.233 kg is moving toward the wall at v = 18.5 m/s and at an angle of θ = 26° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.07 s.

What is the magnitude of the average force the wall exerts on the racquet ball?

Answer 1

Force is the rate of change of momentum.
So average force = change in momentum / time
Momentum before hitting the wall
\[p1 = m\ v\ cos \theta \ \hat{x} - m\ v\ sin\ \theta\ \hat{y}\]
Momentum after hitting the wall
\[p2 = -m\ v\ cos \theta \ \hat{x} - m\ v\ sin\ \theta\ \hat{y}\]
Change in momentum = p2-p1
So
\[p2 - p1 = -2 \ m\ v\ cos \theta \ \hat{x}\]
So, force = (p2-p1)/ t
Put the values of m,v , theta and t and calculate average force. it is in - x direction.