Question

A particle P moving with speed $v$ undergoes a head-on elastic collision with another particle $Q$ of identical mass but at rest. After the collision

• A. $P$ and $Q$ move in opposite directions with speed $\frac{v}{\sqrt{2}}$
• B. Both $P$ and $Q$ move forward with speed $\frac{v}{\sqrt{2}}$
• C. $P$ comes to rest and $Q$ moves forward with speed $v$
• D. Both $P$ and $Q$ move forward with speed $\frac{v}{2}$

Answer 1

The correct option is C $P$ comes to rest and $Q$ moves forward with speed $v$

Hint: “Head-on elastic collision”

Formula used:

$P_i=P_f$

$e=\frac{v_2-v_1}{u_1-u_2}$

Solution:

Let us assume that $u_1$ and $u_2$are the velocities before collision.

Let us assume that $v_1$ and $v_2$ are the velocities after collision

$p_i=p_f$

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\dots \left(i\right)$
As
$e=\frac{v_2-v_1}{u_1-u_2}\dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$
$v_1=\left(\frac{m_1-e{m}_2}{m_1+m_2}\right)u_1+\left(\frac{(1+e)m_2}{m_1+m_2}\right)u_2$

$v_2=\left(\frac{m_2-e{m}_1}{m_1+m_2}\right)u_2+\left(\frac{(1+e)m_1}{m_1+m_2}\right)u_1$

Velocity gets exchanged as $u_1=v{u}_2=0$

$v_1=0$

$v_2=u_1=v$

Final Answer : $\left(c\right)$