A steel ball of mass m = 1 kg and a cord of negligible mass and length L = 2 m m

Question

A steel ball of mass m = 1 kg and a cord of negligible mass and length L = 2 m make up a simple pendulum that can pivot without friction about the point O (see 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 m = 1 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and take the coefficient of kinetic friction between the block and shelf to be 0.1.

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

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

Explanation / Answer

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

0.5mv² = mgh
v = [2gh]
= [2(9.81m/s²)(2m)]
= 6.26 m/s

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

0 = mv + mv
v = - mv/m
= -(1.0kg)(6.26m/s) / 1.0kg
= -6.26m/s

B. The acceleration of the block is:

F = ma = -µmg
a = -µg
= -0.1(9.81m/s²)
= -9.81m/s²

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

v² = v² + 2ax
x = (v² - v²) / 2a
= [0 - (-6.26m/s)²] / 2(-9.81m/s²)
= 1.997m

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