Question

A bomb burst in three fragments in which two fragments of equal mass of $1kg$ each moving with speed $v$ right angle to each other while third fragment of having mass twice that of either masses. If kinetic energy of system (all masses) increase by $96J$ after bursting then find out CORRECT statement about third fragment:

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

Answer 1

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

From conservation of momentum

Resultant momentum of first two masses

$P=1\times v\sqrt{2}$ is along angle bisector of the two velocities.

Momentum of third mass

$P=-v\sqrt{2}$ in opposite direction of the resultant of previous two momentum.

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

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

Now total kinetic energy

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

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

$v=8m/s$

Thus velocities of first two masses are $8m/s$ each and perpendicular to each other. While velocity of third mass

$v_3=\frac{8}{\sqrt{2}}$

$v=4\sqrt{2}m/s$

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