Question

Particles of mass $m$ and $M$ moving with horizontal veocity $u_1=4\text{m/sec}$ and $u_2=6\text{m/s}$ as shown in figure. If $m<<M$ then for one dimensional elastic collision, the velocity of lighter particle after collision will be

• A. $4\text{m/s}$ opposite to its original direction.
• B. $16\text{m/s}$ along its original direction.
• C. $12\text{m/s}$ opposite to its original direction.
• D. $16\text{m/s}$ opposite to its original direction.

Answer 1

The correct option is D $16\text{m/s}$ opposite to its original direction.

Before collision

After collision


From the concept of collision we know
$v_1=\left(\frac{m-M}{m+M}\right)u_1+\frac{2M{u}_2}{m+M}$
$[\text{As}m<<M]$
$\Rightarrow v_1=-u_1+2u_2=(-4+2\times (-6\left)\right)\text{m/s}=-16\text{m/s}$
Here $-ve$ indicates ball will have final velocity of $16\text{m/s}$ opposite to its original direction.