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
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
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