Question

A blast breaks a body initially at rest of mass $0.5\text{kg}$ into three pieces, two smaller pieces of equal mass and the third double the mass of either of small piece. After the blast the two smaller masses move at right angles to one another with equal speed. Find the statements that is/are true for this case assuming that the energy of blast is totally transferred to masses.

• A. All the three pieces share the energy of blast equally
• B. The speed of bigger mass is $\sqrt{2}$ times the speed of either of the smaller mass
• C. The direction of motion of bigger mass makes an angle of ${135}^{\circ }$ with the direction of smaller pieces
• D. The bigger piece carries double the energy of either piece.

Answer 1

Answer: (A), (C)

The correct options are
A All the three pieces share the energy of blast equally
C The direction of motion of bigger mass makes an angle of ${135}^{\circ }$ with the direction of smaller pieces


Total mass of the body is $0.5\text{kg}$, hence the masses of the broken parts will be $m,m$ and $2m$ respectively.
Let smaller masses have velocity $v$.
${\overrightarrow{P}}_{initial}=0$
By conservation of momentum,
$0={\overrightarrow{P}}_1+{\overrightarrow{P}}_2+{\overrightarrow{P}}_3$
$\Rightarrow 0=-mv\hat{i}-mv\hat{j}+\left(2m\right){\overrightarrow{v}}_3$
$\Rightarrow {\overrightarrow{v}}_3=\frac{1}{2}v(\hat{i}+\hat{j})$

$\Rightarrow |{\overrightarrow{v}}_3|=\frac{v}{\sqrt{2}}$
Final kinetic energy of all the pieces are $\frac{1}{2}m{v}^2,\frac{1}{2}m{v}^2$ and $\frac{1}{2}\left(2m\right){\left(\frac{v}{\sqrt{2}}\right)}^2=\frac{1}{2}m{v}^2$
Hence, all the three pieces share the blast energy equally.

Again, $v_3=\frac{v}{\sqrt{2}}=v\sin \theta$
$\Rightarrow \theta ={45}^{\circ }$
Hence, angle between bigger mass $2m$ and each of the smaller mass is ${135}^{\circ }$.