Question

A 4.3 kg steel ball and 6.5 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 4.3 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 0.9. The acceleration of gravity is 9.81 m/s².

Answer 1

given,

mass of steel ball, M = 4.3 kg

length of the chord, L = 6.5 m

mass of the block, m = 4.3 Kg

coefficient of friction, μ = 0.9

acceleration due to gravity, g = 9.81 m/s²

here the potential energy of the bob is converted into kinetic energy

`m g L = (1)/(2) mv^2`

`v= √(2gL)`

`v= √(2* 9.8* 6.5)`

v = 11.29 m/s

As the collision is elastic the velocity of the block is same as that of bob.

now,

work done by the friction force = kinetic energy of the block

`f . d = (1)/(2) mv^2`

`\mu m g. d = (1)/(2) mv^2`

`d=(v^2)/(2\mu g)`

`d=(11.29^2)/(2* 0.9 * 9.8)`

d = 7.23 m

the distance traveled by the block will be equal to 7.23 m.