A steel ball of mass m 1 = 1.1 kg and a cord of length of L = 2.2 m of negligibl

Question

A steel ball of mass m1 = 1.1 kg and a cord of length of L = 2.2 m of negligible mass make up a simple pendulum that can pivot without friction about the point O, as in the figure below. This pendulum is released from rest in a horizontal position, and when the ball is at its lowest point it strikes a block of mass m2 = 1.1 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and that the coefficient of kinetic friction between the block and shelf is 0.10.

(a) What is the velocity of the block just after impact?
m/s

(b) How far does the block slide before coming to rest (assuming that the shelf is long enough)?
m

Explanation / Answer

A)

The speed of the ball at its lowest point is:

0.5mv² = mgh
v = [2gh]
= [2(9.81m/s²)(2.2m)]
= 6.569931506m/s

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

0 = mv + mv
v = - mv/m
= -(1.1kg)(6.569931506m/s) / 1.1kg
= -6.569931506m/s

B)

The acceleration of the block is:

F = ma = -µmg
a = -µg
= -0.10(9.81m/s²)
= -0.981m/s²

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

v² = v² + 2ax
x = (v² - v²) / 2a
= [0 - (-6.569931506m/s)²] / 2(-0.981m/s²)
= 3.348588943m

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