Diagram: http://oi42.tinypic.com/212beyc.jpg A 3.6 kg steel ball and 1 m cord of

Question

Diagram: http://oi42.tinypic.com/212beyc.jpg

A 3.6 kg steel ball and 1 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point O. This pendulum is released from rest in a horizontal position and when the ball is at its lowest point it strikes a 1.4 kg block 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 0.98.


How far does the block move before coming to rest?
The acceleration due to gravity is 9.81 m/s2

Explanation / Answer

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

0.5mv² = mgh
v = [2gh]
= [2(9.81m/s²)(1m)]
= 4.43m/s

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

0 = mv + mv
v = - mv/m
= -(3.6kg)(4.43m/s) / 1.4kg
= -11.4m/s

B. The acceleration of the block is:

F = ma = -µmg
a = -µg
= -0.98(9.81m/s²)
= -9.61m/s²

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

v² = v² + 2ax
x = (v² - v²) / 2a
= [0 - (-4.43m/s)²] / 2(-9.61m/s²)
= 1.02m

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