Question

6. A 7-kg bowling ball moving at 6.0 m/s strikes a 1-kg bowling pin. 11 we van

Answer 1

Answer:

210 N

Explanation:

The change in momentum of the ball is equal (in magnitude) to the impulse given to the pin, according to

$\Delta p = F \Delta t$ (1)

where

$\Delta p$ is the change in momentum of the ball

F is the force exerted on the pin

$\Delta t$ is the duration of the collision

For the bowling ball, we have

m = 7 kg (mass)

u = 6.0 m/s (initial velocity)

v = 4.5 m/s (final velocity)

So, the change in momentum (in magnitude) is

$\Delta p = m(u-v)=(7)(6.0-4.5)=10.5 kg m/s$

We also know the duration of the collision,

$\Delta t = 0.05 s$

And so if we re-arrange eq.(1), we find the force exerted on the pin:

$F=\frac{\Delta p}{\Delta t}=\frac{10.5}{0.05}=210 N$