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

Answer

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

$m g L = \dfrac{1}{2} mv^2$

$v= \sqrt{2gL}$

$v= \sqrt{2\times 9.8\times 6.5}$

      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

$f . d = \dfrac{1}{2} mv^2$

$\mu m g. d = \dfrac{1}{2} mv^2$

$d=\dfrac{v^2}{2\mu g}$

$d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}$

    d = 7.23 m

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