Question

1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is in contact with the wall for 0.0210 seconds, what is the force exerted
on the ball by the wall?
O 1050 N
22.1 N
0 402 N
0 651 N
CAN SOMEONE PLS PLS ACTUALLY HELP ME

Answer 1

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N