Question

A body of mass $1kg$ initially at rest, explodes and breaks into three fragments of mass in the ratio $1:1:3$. The two pieces of equal mass fly off perpendicular to each other with speed of $15m/s$. each. The speed of heavier fragment is:

• A. $7m/s$
  
• B. $45m/s$
  
• C. $5m/s$
  
• D. $156m/s$
  

Answer 1

Answer: (A)

$7m/s$

Total mass of the body $M=1kg$

So, mass of first fragment $m_1=\frac{1}{5}\times 1=0.2kg$

Mass of second fragment $m_2=\frac{1}{5}\times 1=0.2kg$

Mass of third fragment $m_3=\frac{3}{5}\times 1=0.6kg$

Velocity of masses $m_1$ and $m_2$, $v=15m/s$

Momentum of first fragment $p_1=m_{1}v=0.2\times 15=3kgm/s$

Similarly, $p_2=m_{2}v=0.2\times 15=3kgm/s$

Third fragment will fly in a direction opposite to the direction of resultant of $p_1$ and $p_2$ so that total momentum of the system after explosion remains zero.

Momentum of the third fragment $p_3=\sqrt{p_1^2+p_2^2}$

$\therefore$ $p_3=\sqrt{3^2+3^2}=3\sqrt{2}kgm/s$

Thus speed of third fragment $v_3=\frac{p_3}{m_3}=\frac{3\times 1.414}{0.6}=7.07m/s$