Question

A bullet of mass $A$ and velocity $B$ is fired into a block of wood of mass $C.$ If loss of any mass and friction be neglected, then velocity of the system will be
(A) $\dfrac{{AB}}{{A + C}}$
(B) $\dfrac{{A + C}}{{B + C}}$
(C) $\dfrac{{AC}}{{B + C}}$
(D) $\dfrac{{A + B}}{{AC}}$

Answer 1

Calculate the momentum of the system before the impact and after the impact. Then use the law of conservation of momentum to solve the question.

Complete step by step answer: It is given in the question that
the mass of the bullet is $A$.
Velocity of bullet is $B$.
Mass of block is $C$
Velocity of the block is zero as it is not moving.
According to the law of conservation of momentum we know that the momentum before the impact is equal to the momentum after the impact.
i.e. ${p_i} = {p_a}$ . . . (1)
Where,
${p_i}$is the momentum before the impact.
${p_a}$is the momentum after the impact.
Now, momentum is the product of mass and velocity.
Therefore, momentum the system before the impact
${p_i} = AB + C \times 0$
And momentum of the system after the impact is.
${p_a} = (A + C){V_s}$
Where,
${V_s}$ is the velocity of the system.
By substituting the values of ${p_i}$ and ${p_a}$ in equation (1), we get
$AB = (A + C){V_s}$
By dividing both the sides by $(A + C)$, we get
${V_s} = \dfrac{{AB}}{{A + C}}$
Hence , the velocity of the system will be $\dfrac{{AB}}{{A + C}}$.

Note: Before the impact, the bullet and block are separate. Therefore, the momentum of the bullet and the momentum of the block are calculated separately and then added. But after the impact, the bullet and the block are together therefore the mass of the bullet and block will be added first and then it will be multiplied with the velocity of the system.