Question

A bullet of mass \[1 \times {10^{ - 2}}Kg\] moving horizontally with a velocity of \[2 \times {10^2}m{s^{ - 1}}\]strikes a block of mass \[1.99Kg\]and gets embedded into it. If the coefficient of kinetic friction between the block and the horizontal surface to\[0.25\], then the distance moved by the block with the bullet before coming to rest is:
(A) $0.4m$
(B) $0.8m$
(C) $0.6m$
(D) $0.2m$

Answer 1

Hint: In order to find the distance covered by the block we need to use the theorem of conservation of momentum. This will give us the velocity with which the block moves. Then using the equation of motions and taking the role of friction into consideration, we can compute the distance travelled by the block.

Complete Step-By-Step Solution:
Applying conservation of momentum, we know:
In this case, the momentum of the block after being hit by the bullet should be the same as the momentum of the bullet hitting it.
We know that the quantity momentum is a product of mass and velocity.
According to the conservation of momentum, mathematically we can write:
${p_1}{v_1} = {p_2}{v_2}$
Where:
${p_1}{v_1}$ is the initial momentum and ${p_2}{v_2}$ is the final momentum of the system.
In this case, let us consider ${m_{}}$ be the mass of the bullet and $M$be the combined mass of bullet and block.
Let $v$ be the velocity of the bullet and $V$be the velocity of the block after being struck by the bullet.
Thus applying the conservation of momentum:
$mv = MV$
Applying the values as given in the question:
$0.01 \times 200 = \left( {0.01 + 1.99} \right) \times V$
On solving the equation, we get:
$V = 1m/s$
Given, the coefficient of friction is $0.25$
We know, frictional force can be calculated as:
$F = \mu (m + M)g$
Putting the values, we obtain:
$0.25(0.01 + 1.99)g$
Now, from Newton's second law of motion, we know, force is a product of mass and acceleration.
Hence, we can write:
$(0.01 + 1.99)a = 0.25(0.01 + 1.99)g$
On solving the equation, we get, $a = 2.5m/{s^2}$
Now, applying the obtained values in the equation of motion, ${V^2} - {u^2} = 2aS$
We obtain:
$S = \dfrac{{{1^2}}}{{(2 \times 2.5)}}$
Hence, on solving, we get:
$S = 0.2m$

Thus, option (D) is correct.

Note:
We know that the theorem of conservation of momentum states that the total momentum of a system is conserved. This means that the initial momentum and the final momentum of the system remains constant.