Question

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

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

Answer 1

Answer: (A), (C)

The direction of motion of the bigger mass makes an angle of ${135}^{\circ }$ with the direction of the smaller pieces.

Given that mass of the body initially at rest is $0.5\text{kg}$.


Also given, after the blast
$m_1=m_2=m$ and $m_3=2m$
$v_1=v_2=v$

Let velocity of $m_1$ along $x-$axis $\overrightarrow{v_1}=v\hat{i}$

So, velocity of $m_2$ along $y-$axis $\overrightarrow{v_2}=v\hat{j}$

Since there is no external force acting on the body, so momentum remains conserved.

Conservation of linear momentum gives

$\overrightarrow{p_i}=\overrightarrow{p_f}$

$\Rightarrow (m_1+m_2+m_3)\times 0=m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}$

$\Rightarrow 0=mv\hat{i}+mv\hat{j}+2m\overrightarrow{v_3}$

$\Rightarrow \overrightarrow{v_3}=-\frac{v}{2}(\hat{i}+\hat{j})$

$\Rightarrow |\overrightarrow{v_3}|=\frac{v}{\sqrt{2}}$

Now, writing energies of the particles after explosion

$E_1=E_2=\frac{1}{2}m{v}^2$

and, $E_3=\frac{1}{2}\left(2m\right)\times v_3^2$

$\Rightarrow E_3=\frac{1}{2}\left(2m\right)\times \frac{v^2}{2}=\frac{1}{2}m{v}^2$

$\Rightarrow E_1=E_2=E_3$

Now, the angle between $\overrightarrow{v_1}$ and $\overrightarrow{v_3}$ can be found with the help of dot product.

$\overrightarrow{v_1}\cdot \overrightarrow{v_3}=|\overrightarrow{v_1}||\overrightarrow{v_3}|\cos \theta$

$\Rightarrow v\hat{i}\cdot (-\frac{v}{2}(\hat{i}+\hat{j}\left)\right)=\left(v\right)\frac{v}{\sqrt{2}}\cos \theta$

$\Rightarrow \frac{-v^2}{2}=\frac{v^2}{\sqrt{2}}\cos \theta$

$\Rightarrow \theta ={\cos }^{-1}(-\frac{1}{\sqrt{2}})={135}^{\circ }$

Angle between $\overrightarrow{v_2}$ and $\overrightarrow{v_3}={360}^{\circ }-{90}^{\circ }-{135}^{\circ }={135}^{\circ }$

Hence, velocity of the bigger mass makes an angle of ${135}^{\circ }$ with the direction of the smaller pieces.

Therefore, the correct options are $\left(a\right)$ and $\left(c\right)$.