Question

A bullet of mass $0.01kg$ and traveling at a speed of $500m/s$ strikes and passes horizontally through a block of mass $2kg$ which is suspended by a string of length $5m$. The center of gravity of the block is found to raise a vertical distance of $0.1m$. What is the speed of the bullet after it emerges from the block ($g=9.8m/s^2$) (time of the passing of bullet is negligible)

Answer 1

Given: A bullet of mass 0.01kg and traveling at a speed of 500m/s strikes and passes horizontally through a block of mass 2kg which is suspended by a string of length 5m. The center of gravity of the block is found to raise a vertical distance of 0.1m.

To find the speed of the bullet after it emerges from the block

Solution:

Suppose $v_1$ and $v_2$ are the velocities of the bullet and the block after collision.

Since the block rises to a height of $h=0.1m$, so all its kinetic energy is converted into its potential enegy

Thus by conservation of energy

$\frac{1}{2}{m}_2{v}_2^2=m_{2}gh⟹v_2=\sqrt{2gh}$

Substituting the corresponding values we get

$v_2=\sqrt{2\times 9.8\times 0.1}⟹v_2=1.4m/s$

If $u_1$ is the initial velocity of the bullet, then applying the law of conservation of momentum along the initial direction of the bullet, we get

$m_1{u}_1=m_1{v}_1+m_2{v}_2⟹v_1=\frac{m_1{u}_1-m_2{v}_2}{m_1}$

Now substituting the values given , we get

$⟹v_1=\frac{0.01\times 500-2\times 1.4}{0.01}⟹v_1=220m/s$

is the speed of the bullet after it emerges from the block