A 4 . 3 kg steel ball and 4 . 5 m cord of negli- gible mass make up a simple pen

Question

A 4.3 kg steel ball and 4.5 m cord of negli- gible mass make up a simple pendulum that can pivot without friction about the point O. This pendulum is released from rest in a hor- izontal position and when the ball is at its lowest point it strikes a 4.3 kg block sitting at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of friction between the block and shelf to be 0.54.

The acceleration of gravity is 9.81 m/s2.

What is the velocity of the block just after impact?

Answer in units of m/s.

Part B:

How far does the block move before coming to rest?

Answer in units of m.

Explanation / Answer

A. The speed of the ball at its lowest point is:

0.5mv² = mgh
v = [2gh]
= [2(9.81m/s²)(4.5m)]
= 9.396 m/s

Then, using the law of conservation of momentum, where the ball is m, the block m:

0 = mv + mv
v = - mv/m
= -(4.3kg)(9.396m/s) / 4.3kg
= -4.3 m/s

B. The acceleration of the block is:

F = ma = -µmg
a = -µg
= -0.54(9.81m/s²)
= -5.297 m/s²

So, it will slide to a stop in a distance of:

v² = v² + 2ax
x = (v² - v²) / 2a
= [0 - (-9.396m/s)²] / 2(-5.297 m/s²)
= 1.25 m

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