Question

A 0.21 kg baseball moving at +25 m/s is slowed to a stop by a catcher who exerts a constant force of −360 N. a) How long does it take this force to stop the ball? Answer in units of s.

Answer 1

Answer:

0.0146 s

Explanation:

First of all, we can find the acceleration of the ball by using Newton's second law of motion:

$F=ma$

where:

F = -360 N is the force acting on the ball

m = 0.21 kg is the mass of the ball

a is the acceleration

Solving for a,

$a=\frac{F}{m}=\frac{-360}{0.21}=-1714 m/s^2$

Then the ball moves by uniformly accelerated motion, so we can use the following suvat equation:

$v=u+at$

where:

u = +25 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1714 m/s^2$ is the acceleration

t is the time it takes for the ball to stop

Solving for t,

$t=\frac{v-u}{a}=\frac{0-(25)}{-1714}=0.0146 s$