Question

A 0.1 kg ball hits a wall at right angles to the surface and bounces off. Before the collision, the ball’s velocity component perpendicular to the wall is vi,x = −8 m/s. After the collision, it is vf,x = +6m/s. What is ∆px, the change in the ball’s momentum perpendicular to the wall?

A. −1.4kgm/s

B. +1.4kgm/s

C. +0.8kgm/s

D. −0.6kgm/s

E. −0.2kgm/s

Answer 1

Here,

m = 0.1 Kg

initial velocity, vi = - 8 m/s

final velocity, vf = 6 m/s

for the change in momentum

change in momentum, ∆px = final momentum - initial momentum

change in momentum, ∆px = 0.10 * (6 - (-8))

change in momentum, ∆px = + 1.4 Kg.m/s

the change in momentum, ∆px is B) + 1.4 Kg.m/s