An explosion blows a stationary rock into thr | vedclass

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$(A)$ Given: $m_1 = 1 \,kg, m_2 = 2 \,kg$, $v_1 = 12 \,ms^{-1}, v_2 = 8 \,ms^{-1}$, and $v_3 = 4 \,ms^{-1}$.Since the rock is initially stationary, th

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$8$

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$(A)$ Given: $m_1 = 1 \,kg, m_2 = 2 \,kg$, $v_1 = 12 \,ms^{-1}, v_2 = 8 \,ms^{-1}$, and $v_3 = 4 \,ms^{-1}$.Since the rock is initially stationary, the initial momentum is zero. By the law of conservation of linear momentum, the final momentum must also be zero:$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \Rightarrow \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$.The magnitude of the momentum of the third part is given by:$p_3 = \sqrt{p_1^2 + p_2^2 + 2p_1 p_2 \cos 90^{\circ}} = \sqrt{(m_1 v_1)^2 + (m_2 v_2)^2}$.Substituting the values:$p_3 = \sqrt{(1 	imes 12)^2 + (2 	imes 8)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \,kg \cdot ms^{-1}$.Since $p_3 = m_3 v_3$, we have $m_3 	imes 4 = 20$, which gives $m_3 = 5 \,kg$.The total mass of the rock is $m = m_1 + m_2 + m_3 = 1 + 2 + 5 = 8 \,kg$.Thus, the correct option is $A$.

An explosion blows a stationary rock into three parts. Two parts of masses $1 \,kg$ and $2 \,kg$ move at right angles to one another with velocities $12 \,ms^{-1}$ and $8 \,ms^{-1}$, respectively. If the velocity of the third part is $4 \,ms^{-1}$, the mass of the rock is (in $\,kg$)

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