Question

A bullet of mass $A$ and velocity $B$ is fired into a wooden block of mass $C$. If the bullet enclosed in the wooden block, then the magnitude of velocity of the system just after the collision will be
(A) $\dfrac{{A + B}}{{AC}}$
(B) $\dfrac{{A + C}}{{B + C}}$
(C) $\dfrac{{AC}}{{B + C}}$
(D) $\dfrac{{AB}}{{A + C}}$

Answer 1

From the principle of conservation of momentum, the total momentum before collision is equal to the total momentum after collision. Since we speak about the velocity of the system, the velocity of the two objects after the collision are the same.

Formula used: In this solution we will be using the following formulae;
$p = mv$ where $p$ is the momentum of a body, $m$is its mass and $v$ is the velocity of the body.
$\sum {{p_{ii}} = \sum {{p_{io}}} }$ where ${p_{ii}}$is the individual momentum of the objects before a collision, and ${p_{io}}$ is the individual momentum of this same object after momentum.

Complete Step-by-Step Solution:
A bullet of mass A and velocity B is said to be fired into a wooden block of mass C. we are to find the velocity of the system.
To do so, we must use the principle of conservation of momentum which states that the total momentum before a collision is equal to the total momentum after collision. Mathematically,
$\sum {{p_{ii}} = \sum {{p_{if}}} }$ where ${p_{ii}}$is the individual momentum of the objects before a collision, and ${p_{io}}$ is the individual momentum of this same object after momentum.
Hence, for the situation in question
${m_b}{v_{ib}} + {m_w}{v_{iw}} = {m_b}{v_{bf}} + {m_w}{v_{wf}}$ since momentum is $p = mv$ where $m$is its mass and $v$ is the velocity of the body.
Now, since the velocity of the system after collision is of concern, we have
${v_{wf}} = {v_{bf}} = v$
From question, the mass of the bullet is A, and velocity is B but wooden block is static i.e. ${m_b} = A,{v_{ib}} = B$ and ${v_{iw}} = 0$
Hence,
$AB + 0 = Av + Cv$
$\Rightarrow AB = v(A + C)$
Hence, by dividing by $\Rightarrow AB = v(A + C)$
$v = \dfrac{{AB}}{{A + C}}$

The correct option is hence, D

Note: For clarity, the velocity of the system after collision doesn’t necessarily have to mean that the two objects move with the same velocity.