Question

A particle of mass m moving in x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
(A) 44%
(B) 50%
(C) 56%
(D) 62%

Answer 1

Use the law of conservation of energy to determine the final velocity of the system. The percentage loss in the kinetic energy of the system during the collision is the ratio of difference in the final and initial kinetic energy and the initial kinetic energy of the system.

Formula used:
${m_1}{\vec u_1} + {m_2}{\vec u_2} = \left( {{m_1} + {m_2}} \right)\vec V$
Here, ${m_1}$ is the mass of the first particle, ${m_2}$ is the mass of the second particle, ${\vec u_1}$ is the initial velocity and direction of the first particle, ${\vec u_2}$ is the initial velocity of the second particle and $\vec V$ is the velocity of the final system.

Complete step by step answer:
In inelastic collision, the linear momentum of the system is conserved but the kinetic energy of the system does not conserve.
According to the law of conservation of linear momentum in an inelastic collision, we can write,
${m_1}{\vec u_1} + {m_2}{\vec u_2} = \left( {{m_1} + {m_2}} \right)\vec V$
Here, ${m_1}$ is the mass of the first particle, ${m_2}$ is the mass of the second particle, ${\vec u_1}$ is the initial velocity and direction of the first particle, ${\vec u_2}$ is the initial velocity of the second particle and $\vec V$ is the velocity of the final system.
The first particle has mass m and the second particle has mass 2m. Also, the initial velocity of the first particle is 2v in x-direction and that of the second particle is v in y-direction. therefore, the above expression becomes,
$m\left( {2v\hat i} \right) + \left( {2m} \right)\left( {v\hat j} \right) = \left( {m + 2m} \right)\vec V$
$\Rightarrow 2mv\left( {\hat i + \hat j} \right) = 3m\vec V$
$\vec V = \dfrac{2}{3}v\left( {\hat i + \hat j} \right)$
Therefore, the magnitude of the final velocity of the system is,
$\left| {\vec V} \right| = \dfrac{2}{3}v\sqrt {{1^2} + {1^2}}$
$\Rightarrow \left| {\vec V} \right| = \dfrac{{2\sqrt 2 }}{3}v$
Therefore, the final kinetic energy of the system is,
${K_f} = \dfrac{1}{2}m{\left| {\hat V} \right|^2}$
Substitute $\left| {\vec V} \right| = \dfrac{{2\sqrt 2 }}{3}v$ in the above equation.
${K_f} = \dfrac{1}{2}m{\left( {\dfrac{{2\sqrt 2 }}{3}v} \right)^2}$
$\therefore {K_f} = \dfrac{4}{3}m{v^2}$
The initial kinetic energy of the system is,
${K_i} = \dfrac{1}{2}\left( m \right){\left( {2v} \right)^2} + \dfrac{1}{2}\left( {2m} \right){\left( v \right)^2}$
$\Rightarrow {K_i} = 3m{v^2}$
Now, the percentage loss in the energy during the collision is,
$E\left( \% \right) = \dfrac{{{K_i} - {K_f}}}{{{K_i}}}$
Substitute ${K_f} = \dfrac{4}{3}m{v^2}$ and ${K_i} = 3m{v^2}$ in the above equation.
$E\left( \% \right) = \dfrac{{3m{v^2} - \dfrac{4}{3}m{v^2}}}{{3m{v^2}}}$
$\Rightarrow E\left( \% \right) = \dfrac{{\dfrac{5}{3}m{v^2}}}{{3m{v^2}}}$
$\Rightarrow E\left( \% \right) = 0.56 = 56\%$
So, the correct answer is option (C).

Note: Always consider the direction of the velocity of the particle. Here, the first particle is along the x-axis and the second particle is along the y-axis. The net direction of these two particles is the direction of the final system.