Question

A bullet of mass 10 g moving horizontally with a velocity of $400\text{ m/s}$ strikes a wooden block of mass $2kg$ which is suspended by a light inextensible string of $5\text{ m}$. As a result, the center of gravity of the block is found to rise a distance of $10\text{ cm}$. The speed of the bullet after it emerges horizontally out of the block is:
A. $120\text{ m/s}$
B. $160\text{ m/s}$
C. $100\text{ m/s}$
D. $80\text{ m/s}$

Answer 1

Hint: The bullet undergoes a collision with the wooden block such that it transfers energy from the bullet to the wooden block. The energy gained by the black is used to increase the potential energy of the wooden block. The kinetic energy of the bullet is reduced due to the collision. Also, the momentum is conserved in all these situations.

Complete answer:
In the problem, we have a bullet of mass 10 g travelling with a velocity of $400\text{ m/s}$. This bullet hits a block of mass $2kg$, as a result, the block gains velocity and the velocity of the bullet will be reduced due to the transfer of kinetic energy to the block and due to other dissipative forces. The bullet-block is an example of an inelastic collision in which the only conserved quantity is the momentum. So, according to the conservation of momentum, initial momentum should be equal to the final momentum. So, we can write,
${{m}_{1}}{{u}_{1}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$
Where,
${{m}_{1}}$ is the mass of the bullet.
${{m}_{2}}$ is the mass of the wooden block.
${{u}_{1}}$ is the initial velocity of the bullet.
${{v}_{1}}$ is the final velocity of the bullet.
${{v}_{2}}$ is the final velocity of the block.
Since the block is initially at rest, its initial momentum is zero. That is why you don’t see an initial momentum term for the block.
Substituting the values in the above equation, we get,
$\left( 10g \right)\times \left( 400m{{s}^{-1}} \right)=\left( 10g \right)\times {{v}_{1}}+\left( 2kg \right)\times {{v}_{2}}$ … equation (1)
The velocity of the block can be found out from the energy conservation theorem. The block gains potential energy after the collision with the bullet, this potential energy is converted into the kinetic energy when the block comes down from the height. So, we can write,
$mgh=\dfrac{1}{2}m{{v}_{2}}^{2}$
$\Rightarrow {{v}_{2}}=\sqrt{2gh}$
${{v}_{2}}=\sqrt{2\times 9.8\times 0.1}$
$\therefore {{v}_{2}}=1.4m{{s}^{-1}}$ … equation (2)
Substituting equation (2) in equation (1), we get,
$\left( 10g \right)\times \left( 400m{{s}^{-1}} \right)=\left( 10g \right)\times {{v}_{1}}+\left( 2kg \right)\times \left( 1.4m{{s}^{-1}} \right)$
$\Rightarrow \left( 4kgm{{s}^{-1}} \right)=\left( 10g \right)\times {{v}_{1}}+\left( 2.8kgm{{s}^{-1}} \right)$
$\Rightarrow {{v}_{1}}=\dfrac{\left( 4kgm{{s}^{-1}} \right)-\left( 2.8kgm{{s}^{-1}} \right)}{\left( {{10}^{-2}}kg \right)}$
$\therefore {{v}_{1}}=120m{{s}^{-1}}$
So, the velocity of the bullet after the collision is ${{v}_{1}}=120m{{s}^{-1}}$.
So, the answer to the question is option (A).

Note: The coefficient of restitution can be defined as the ratio of relative velocity after the collision to the relative velocity before the collision.
For an inelastic collision, the coefficient of restitution is always less than one and greater than zero $\left( 0If the value of e is less than zero or negative, it means that the two objects continue to move in the same direction before and after the collision.
Coefficient of restitution can also be defined as the ratio of the square root of the final kinetic energy after the collision to the square root of the initial kinetic energy before collision.