Question

If a boy (m = 25 kg) at rest on skates is pushed by another boy who exerts a force of 500 N on him and if the first boy's final velocity is 20 m/s, what was the contact time?

Answer 1

Answer:

1 second

Explanation:

The impulse exerted on the boy is equal to its change of momentum:

$I = \Delta p\\F \Delta t = m \Delta v$

where

F = 500 N is the push on the boy

$\Delta t$ is the contact time

m = 25 kg is the mass of the boy

$\Delta v = 20 m/s$ is the change in velocity of the boy

Solving the formula for the contact time, we find

$\Delta t=\frac{m\Delta v}{F}=\frac{(25 kg)(20 m/s)}{500 N}=1 s$