Question

A $6\, kg$ bomb at rest explodes into three equal pieces $P, Q$ and $R$. If $P$ flies with speed $30\, m / s$ and $Q$ with speed $40\, m / s$ making an angle $90^{\circ}$ with the direction of $P$. The angle between the direction of motion of $P$. and $R$ is about

• A. $143^{\circ}$
• B. $127^{\circ}$
• C. $120^{\circ}$
• D. $150^{\circ}$

Answer 1

Answer: (B)

$P_{P}=30(2)=60\, kg\, ms ^{-1}$
$P_{Q}=40(2)=80\, kg \,ms ^{-1}$
$\tan \theta=\frac{60}{80}=3 / 4$
$\theta=37^{\circ}$
So angle between $P$ and $R$ will be
$90^{\circ}+37^{\circ}=127^{\circ}$