Question

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.48 kg and 2.77 kg, and the length of the wire is 1.11 m. Find the velocity of the ball just after the collision.

Answer 1

Answer: Velocity of the ball just after the collision is -1.414 m/s.

Explanation:

As energy is conserved in a reaction so here, energy before collision will be equal to the energy after collision.

$E_{before} = mgh = E_{after} = \frac{1}{2}mv_{o}^{2}$

where,

m = mass

g = gravitational energy = $9.8 m/s^{2}$

h = height or length

$v_{o}$ = initial velocity

Also here, height is the length of wire. Let the height be denoted by 'L'. Therefore,

$\frac{1}{2}mv_{o}^{2} = mgL\\v_{o}^{2} = 2gL\\v_{o} = \sqrt{2gL}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.11 m}\\= 4.66 m/s$

Formula used to calculate velocity after the collision is as follows.

$v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}]$

where,

$v_{f ball}$ = final velocity of ball after collision

$m_{ball}$ = masses of ball

$m_{block}$ = masses of block

Substitute the values into above formula as follows.

$v_{f ball} = v_{o} [\frac{m_{ball} - m_{block}}{m_{ball} + m_{block}}]\\= 4.66 m/s [\frac{1.48 kg - 2.77 kg}{1.48 kg + 2.77 kg}]\\= 4.66 m/s \times (-0.303)\\= -1.414 m/s$

Thus, we can conclude that velocity of the ball just after the collision is -1.414 m/s.