A 4.5 kg steel ball and 9 m cord of negligible mass make up a simple pendulum th

Question

A 4.5 kg steel ball and 9 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 4.5 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 0.64. The acceleration of gravity is 9.81 m/s 2. What is the velocity of the block just after impact?

Explanation / Answer

By conservation of momentum
m u + m u = m v + m u
m (u - v) = m u---------------------------1 since v =0
For elastic collision, kinetic energy is also conserved
m (u² - v²) = m v²
(u + v)*m (u - v) = m v²
(v + u)*m v* = m v ²
(v + u) = v
We can easily find that
v = 2 m u / (m + m)
--------------------------------------...
u = (2gh) = (2*9.8*9) =176.4 m/s
v = 2 m u / (m + m) = 2*4.5*176.4 / 9 = 176.4 m/s
Using v² = 2gs (since the acceleration due to friction is g)
176.4² = 2*0.64*9.81*s
s = 2478 m

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