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4-1.Newton's Laws of Motion
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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
$150^{\circ}$
Solution
(b)
$P_\rho=30(2)=60 \,kg ms ^{-1}$
$P_Q=40(2)=80 \,kg ms ^{-1}$
$\tan \theta=\frac{60}{80}=3 / 4$
$\theta=37^{\circ}$
So angle between $P$ and $R$ will be $90^{\circ}+37^{\circ}=127^{\circ}$
Standard 11
Physics
