Answer to Question #80505 in Mechanics | Relativity for Ankit

Question

A ball having a mass of 0.5 kg is moving towards the east with a speed of 8.0 ms-1.
After being hit by a bat it changes its direction and starts moving towards the north with
a speed of 6.0 ms−1. If the time of impact is 0.1 s, calculate the impulse and average
force acting on the ball.

Answer 1

Ox:
p1x = m*v1x = 4 kg*m/s
p2x = m*v2x = 0 kg*m/s
Oy:
p1y = m*v1y = 0 kg*m/s
p2y = m*v2y = 3 kg*m/s
Δpx = p2x – p1x = -4 kg*m/s
Δpy = p2y – p1y = 3 kg*m/s
Δp = |Δp|=( (Δpx)^2+ (Δpy)^2)^0.5 = 5 kg*m/s
Average force acting on the ball is defined as
Fp = Δp/Δt
Magnitude of average force acting on the ball is
Fp = | Fp | = |Δp|/ Δt
Fp = (5kg*ms^-1)/0.1 s = 50N
Answer:
impulse of the ball is Δp = 5 kg*m/s, average force acting on the ball is Fp = 50N