Question

A particle of mass m, kinetic energy $K$ and linear momentum $p$ collides head on elastically with another particle of mass $2m$ at rest. Match the following (after collision) –

$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A) momentum of first particle}& \text{(P)}\frac{4}{3}p\\ \text{(B) momentum of second particle}& \text{(Q)}\frac{8K}{9}\\ \text{(C) Kinetic energy of first particle}& \text{(R)}-\frac{p}{3}\\ \text{(D) Kinetic energy of second particle}& \left(S\right)\frac{K}{9}\end{array}$

Which of the following option has the correct combination considering column-I and column-II

• A.
  $D\to R$
• B.
  $A\to P$
• C.
  $B\to P$
• D.
  $C\to Q$

Answer 1

The correct option is C

$B\to P$
Let $m_1$ and $m_2$ be the mass of two collision bodies.

$u_1$ and $u_2$ be their velocities before collisions $v_1$ and $v_1$ be their velocities after collisions
In the given situation,
$m_1=m;m_2=2m{u}_1=u_1;u_2=0$
Using conservation of momentum and energy we have equations.

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_1}{m_1+m_2}\right)u_2...\left(1\right)$

$v_2=\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2+\left(\frac{2m_1}{m_1+m_2}\right)u_1...\left(2\right)$
Putting values given in question in equation (1) and (2) ; We get:
$v_1=\frac{-u_1}{3}$ & $v_2=\frac{2}{3}{u}_1$
According to question
$P=m_1{v}_1$ & $K=\frac{1}{2}{m}_1{v}_1^2$
Final momentum of mass $m_1$
$\Rightarrow m_1{v}_1=\frac{-m{u}_1}{3}=-\frac{P}{3}$

Final K.E of mass $m_1;$
$\Rightarrow \frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_1\times \frac{u_1^2}{9}=\frac{K}{9}$

Final Momentum of mass $m_2$
$\Rightarrow m_2{v}_2=2m{v}_2=2m\times \frac{2}{3}{u}_1=\frac{4}{3}P$
Final KE of mass $m_2\Rightarrow \frac{1}{2}{m}_2{v}_2^2\Rightarrow \frac{1}{2}\times 2m\times \left(\frac{4}{9}{u}_1^2\right)\Rightarrow \frac{8K}{9}$

Matching the option we
$(A\to R)(B\to P)(C\to S)(D\to Q)$