Question

A body at rest breaks up into $3$ parts. If $2$ parts having equal masses fly off perpendicularly each after with a velocity of $12m/s$ then the velocity of the third part which has $3$ times mass of each part is

• A. $6\sqrt{2}m/s$ at ${135}^{\circ }$ from each body
  
• B. $4\sqrt{2}m/s$ at an angle of ${45}^{\circ }$ from each body
  
• C. $4\sqrt{2}m/s$ at ${135}^{\circ }$ from each body
  
• D. $24\sqrt{2}m/s$ at an angle of ${135}^{\circ }$ from each body
  

Answer 1

The correct option is C $4\sqrt{2}m/s$ at ${135}^{\circ }$ from each body


Draw a diagram of given situation.

Find the velocity of the third part.
Formula used: $p_i=p_f$
By the momentum conservation in $x$ direction,
$5m\times 0=m\times 12+3mv\cos \theta$
$\Rightarrow v\cos \theta =-4$
By the momentum conservation in $y$ direction,
$5m\times 0=m\times 12+3mv\sin \theta$
$\Rightarrow v\sin \theta =-4$
The third piece will move with net velocity of $x$ and $y$ direction,
$v=\sqrt{v_x^2+v_y^2}$
$v=\sqrt{(-4){}^2+(-4{)}^2}=4\sqrt{2}m/sec$
From the diagram, angle ${135}^{\circ }$ from each part.
Final Answer: $\left(D\right)$