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(B) Since the bomb is initially at rest,the total initial momentum is zero. By the law of conservation of linear momentum,the vector sum of the moment

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$127$

Vedclass · Answer

(B) Since the bomb is initially at rest,the total initial momentum is zero. By the law of conservation of linear momentum,the vector sum of the momenta of the three pieces must be zero: $\vec{p}_P + \vec{p}_Q + \vec{p}_R = 0$,which implies $\vec{p}_R = -(\vec{p}_P + \vec{p}_Q)$.Each piece has a mass $m = 6 \,kg / 3 = 2 \,kg$.The momentum of piece $P$ is $p_P = m v_P = 2 	imes 30 = 60 \,kg \cdot m/s$.The momentum of piece $Q$ is $p_Q = m v_Q = 2 	imes 40 = 80 \,kg \cdot m/s$.Since $P$ and $Q$ are at $90^{\circ}$ to each other,the magnitude of the resultant momentum of $P$ and $Q$ is $p_{PQ} = \sqrt{p_P^2 + p_Q^2} = \sqrt{60^2 + 80^2} = 100 \,kg \cdot m/s$.For the total momentum to be zero,$\vec{p}_R$ must be equal and opposite to $\vec{p}_{PQ}$. Thus,$p_R = 100 \,kg \cdot m/s$.Let $\alpha$ be the angle between $\vec{p}_R$ and the direction opposite to $\vec{p}_P$. Then $	an \alpha = p_Q / p_P = 80 / 60 = 4/3$,so $\alpha = 53^{\circ}$.The angle between $\vec{p}_R$ and $\vec{p}_P$ is $180^{\circ} - 53^{\circ} = 127^{\circ}$.

$A$ $6 \,kg$ 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 (in $^{\circ}$)

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