Question

A body of mass ${m_1}$​ is moving with a velocity $3\,m{s^{ - 1}}$ collides with another body at rest of mass ${m_2}$​. After collision the velocities of the two bodies are $2\,m{s^{ - 1}}$ and $5\,m{s^{ - 1}}$ respectively along the direction of motion of ${m_1}$​. The ratio $\dfrac{{{m_1}}}{{{m_2}}}$​​ is
A. $\dfrac{5}{{12}} \\ $
B. $5 \\ $
C. $\dfrac{1}{5} \\ $
D. $\dfrac{{12}}{5}$

Answer 1

Hint: We will use the rule of conservation of linear momentum to answer this query, and then we will use equations using this principle to calculate the ratio of the two specified masses.

Formula used:
According to the conservation of linear momentum concept, a system's initial and final momentum are always equal.
${P_i} = {P_f}$
where $P = mv$ denotes the momentum of a body.

Complete step by step solution:
According to the question, we have given that initial conditions of two bodies as
${m_1},{u_1} = 3m{s^{ - 1}}$ and ${m_2},{u_2} = 0$.
So total initial momentum of the system is,
${P_i} = {m_1}{u_1} + {m_2}{u_2} \\
\Rightarrow {P_i} = 3{m_1} \to (i) \\ $
Now, after the collision the final condition of two bodies as ${m_1},{v_1} = 2m{s^{ - 1}}$ and ${m_2},{v_2} = 5\,m{s^{ - 1}}$. So final momentum of the system is,
${P_f} = {m_1}{v_1} + {m_2}{v_2} \\
\Rightarrow {P_f} = 2{m_1} + 5{m_2} \to (ii) \\ $
Now, using the principle of law of conservation of momentum, we have ${P_i} = {P_f}$ we get,
$3{m_1} = 2{m_1} + 5{m_2} \\
\Rightarrow {m_1} = 5{m_2} \\
\therefore \dfrac{{{m_1}}}{{{m_2}}} = 5 \\ $
Hence, the correct answer is option B.

Note: It should be remembered that, other than law of conservation of momentum two basic law of conservation are law of conservation of energy where energy remains conserved and other one is law of conservation of angular momentum in rotational dynamics where total angular momentum of the system remains conserved when there is no external torque applied on the body.