Question

Body $A$ of mass $4m$moving with speed $u$ collides with another body $B$ of mass $2m$ at rest. The collision is head on and elastic in nature. After the collision, the fraction of kinetic energy lost by the colliding body $A$ is

• A. $\frac{1}{9}$
• B. $\frac{8}{9}$
• C. $\frac{4}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is B $\frac{8}{9}$


Body 'A' of mass $4m$ moves with speed $u$ and body 'B' of mass $2m$ is at rest.

So, initial linear momentum of system = $4mu+2m\times 0=4mu$

Let final velocity of body 'A' is $v_1$ and body 'B' is $v_2$

Then, final linear momentum of system = $4m{v}_1+2m{v}_2$

As collision is elastic , $e=1$

So, $v_1=\frac{(4m-2m)\times u}{(4m+2m)}+\frac{2\left(2m\right)\times 0}{(4m+2m)}=\frac{2mu}{6m}$

= $\frac{u}{3}$

Similarly, $v_2=\frac{(2m-4m)\times 0}{(4m+2m)}+\frac{2\left(4m\right)u}{(4m+2m)}=\frac{8mu}{6m}=\frac{4u}{3}$

So, initial kinetic energy of body 'A' = $\frac{1}{2}\left(4m\right)u²=2mu²$

Final kinetic energy of body 'A' =$\frac{1}{2}\left(4m\right)\frac{u²}{9}=\frac{2mu²}{9}$

So, change in kinetic energy = final kinetic energy - initial kinetic energy

= $\frac{2mu²}{9}-2mu²$

= $2mu²[\frac{1}{9}-1]$

= $-\frac{16mu²}{9}$ [ Here negative sign indicates that kinetic energy is lost after collision]

So, fractional loss in kinetic energy =

= $\frac{\left(16mu²/9\right)}{\left(2mu²\right)}$

= $\frac{8}{9}$