Question

A particle of mass ${}^{\prime }{m}^{\prime }$ is moving with speed $8\text{m/s}$ on a frictionless surface as shown in figure. If $m<<M$, then for one dimensional elastic collision, the speed of the lighter particle after collision will be

• A. $4\text{m/s}$ in original direction
• B. $4\text{m/s}$ opposite to the original direction
• C. $6\text{m/s}$ in original direction
• D. $6\text{m/s}$ opposite to the original direction.

Answer 1

The correct option is B $4\text{m/s}$ opposite to the original direction

Given, $u_1=8\text{m/s};u_2=2\text{m/s}$
In the case of elastic collision, when $m_1<<m_2$
Speed of lighter particle $\left(v_1\right)=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\frac{2m_2{u}_2}{m_1+m_2}$
$=-u_1+2u_2$
Here, $m<<M$
$\Rightarrow v_1=-8+2\left(2\right)=-4\text{m/s}$
(-ve sign indicates velocity is opposite to original direction)