Question

A bomb bursts into three fragments , in which two fragments of equal masses of $1\text{kg}$ each are moving with speed $v$ at right angles to each other , while third fragment has mass twice that of either masses. If kinetic energy of system $($all masses$)$ increases by $96\text{J}$ after bursting then find out the correct statement about third fragement:

• A. Will move with speed $4\text{m/s}$ at angle of ${135}^{\circ }$ to either
• B. Will move with speed $4\sqrt{2}\text{m/s}$ at angle of ${135}^{\circ }$ to either
• C. Will move with speed $4\sqrt{2}\text{m/s}$ at angle of ${45}^{\circ }$ to either
• D. Will move with speed $4\text{m/s}$ at angle of ${45}^{\circ }$ to either

Answer 1

The correct option is B Will move with speed $4\sqrt{2}\text{m/s}$ at angle of ${135}^{\circ }$ to either

Given,
individual masses of first two fragments, $m_1=m_2=1\text{kg}$

individual velocity of first two fragments $v_1=v_2=v$

mass of third fragments, $m_3=2m=2\times 1=2\text{kg}$

Resultant momentum of first two masses,

$P=\sqrt{(mv{)}^2+(mv{)}^2}=mv\sqrt{2}$ is along angle bisector of the two velocities.

Since the initial velocity of the combined system remains zero, so the initial momentum system also be zero.

From conservation of momentum principle along the given angular bisector.

$m_3{v}_3+P=0$

where, $v_3$ is the velocity of the third fragment.

Substituting the values in the above equation we get,

$2m{v}_3=-mv\sqrt{2}$

$\Rightarrow v_3=-\frac{v}{\sqrt{2}}$

According to the problem, the kinetic energy of the system increases by $96\text{J}$. So

$K.E_f-K.E_i=96\text{J}....\left(1\right)$

where,
Intial kinetic energy of the system, $K.E_i=0$, since the bomb remains stationary before bursting.

Final kinetic energy of the system,$K.E_f=\frac{1}{2}\times 1\times v^2+\frac{1}{2}\times 1\times v^2+\frac{1}{2}\times 2\times {\left(\frac{-v}{\sqrt{2}}\right)}^2$

From $\left(1\right)$,

$\frac{1}{2}\times 1\times v^2+\frac{1}{2}\times 1\times v^2+\frac{1}{2}\times 2\times {\left(\frac{-v}{\sqrt{2}}\right)}^2-0=96$

$\Rightarrow \frac{3v^2}{2}=96$

$\therefore v=8\text{m/s}$

Thus, velocities of first two masses are $v=8\text{m/s}$.

Velocity of third mass $v_3=-\frac{8}{\sqrt{2}}=-4\sqrt{2}\text{m/s}$

Speed of third mass, $|v_3|=4\sqrt{2}\text{m/s}$

The velocity of the third mass makes angle ${135}^{\circ }$ with the velocity of each of the previous two masses.

Hence, option (b) is the correct answer.