Question

A body of mass $m_1$ elastically collide with another stationary body of mass $m_2$
$\begin{array}{cc}\text{Column - I}& \text{Column -II}\\ \text{(a) After the collision, velocity of second}& \left(p\right)>1\\ \text{body is maximum when}\frac{m_1}{m_2}\text{is}& \\ \text{(b) After the collision, the momentum}& \left(q\right)<1\\ \text{of second body is maximum when}\frac{m_1}{m_2}\text{is}& \\ \text{(c) After the collision, K.E. of second}& \left(r\right)=1\\ \text{body is maximum when}\frac{m_1}{m_2}\text{is}& \\ \text{(D) For elastic collision, coefficient of}& \left(s\right)=0\\ \text{restitution is}& \end{array}$

• A. $a-\left(r\right)b-\left(s\right)c-\left(s\right)d-\left(r\right)$
• B. $a-\left(r\right)b-\left(r\right)c-\left(s\right)d-\left(r\right)$
• C. $a-\left(p\right)b-\left(q\right)c-\left(r\right)d-\left(r\right)$
• D. $a-\left(p\right)b-\left(p\right)c-\left(s\right)d-\left(r\right)$

Answer 1

The correct option is C $a-\left(p\right)b-\left(q\right)c-\left(r\right)d-\left(r\right)$

From conservation of linear momentum
$m_{1}u+m_2\times 0=m_1{v}_1+m_2{v}_2..........\left(i\right)$
$1=\frac{v_2-v_1}{u}........\left(ii\right)$
From equal (i) and (ii)
$v_2=\frac{2m_{1}u}{m_1+m_2},v_2=\frac{2u}{1+\frac{m_2}{m_1}}$
$v_2$ is maximum
when $\frac{m_2}{m_1}=0,i.e{m}_1>>m_2$
(or) $\frac{m_1}{m_2}>1$
So $\left(a\right)\to \left(p\right)$
Momentum of second body $P_2=m_2{v}_2=\frac{2m_1{m}_{2}u}{m_1+m_2}$
$P_2$ is maximum, when $\frac{m_1}{m_2}\to 0$
i.e., $m_1<<m_{2}or\frac{m_1}{m_2}<1$
So $\left(b\right)\to \left(q\right)$
After the collision kinetic energy of second body
$K=\frac{1}{2}{m}_2\frac{4m_1^2{u}^2}{(m_1+m_2{)}^2}$
$\frac{dK}{d{m}_1}=0$
For maximum or minima
$m_1=m_2$ i.e. $\frac{m_1}{m_2}=1$
So $\left(c\right)\to \left(r\right)$
For elastic collision coefficient of resitution $e=1$
So $\left(d\right)\to \left(r\right)$