Question

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.

Answer 1

The final velocity of the ball is +9.25 m/s

Explanation:

We can solve the problem by using the impulse theorem, which states that the impulse exerted on the ball (product of the force exerted and the duration of the collision) is equal to the change in momentum of the ball:

$F\Delta t = m(v-u)$

where

F is the force exerted

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem, we have

m = 0.060 kg

u = -32 m/s

$\Delta t = 45 ms = 0.045 s$

F = +55 N

Therefore we can solve for v to find the final velocity:

$v=u+\frac{F\Delta t}{m}=-32 + \frac{(55)(0.045)}{0.060}=+9.25 m/s$

Learn more about change in momentum:

brainly.com/question/9484203

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