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

0.0146 s

Step-by-step 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=(F)/(m)=(-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=(v-u)/(a)=(0-(25))/(-1714)=0.0146 s`