Question

A force in the negative direction of an x-axis is applied for 17ms to a 0.18 kg ball initially moving at 16.0 m/s in the positive direction of the axis. the force varies in magnitude and results in an impulse of magnitude 4.28 ns. what is the velocity of the ball just after the force is applied? indicate the direction of the velocity by its sign.

Answer 1

Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s

But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity

Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s

The velocity of ball after the force is 7.78 m/s in the direction of the force.