(C) Let the velocity of the third fragment be $v'$.According to the law of conservation of linear momentum, since no external force acts on the system

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(C) Let the velocity of the third fragment be $v'$.According to the law of conservation of linear momentum, since no external force acts on the system, the total initial momentum is equal to the total final momentum.Initial momentum $P_i = 5 M \times 0 = 0$.Let the velocity of the first fragment $(M)$ be $\vec{v}_1 = 2v \hat{i}$ and the velocity of the second fragment $(2M)$ be $\vec{v}_2 = -v \hat{i}$ (since they move in opposite directions).Final momentum $P_f = M(2v \hat{i}) + 2M(-v \hat{i}) + 2M(\vec{v}') = 0$.$2Mv \hat{i} - 2Mv \hat{i} + 2M(\vec{v}') = 0$.$0 + 2M(\vec{v}') = 0$.Therefore, $\vec{v}' = 0$.The third fragment will be at rest.

Class 11 Physics Newton's Laws of Motion and Friction Conservation of Linear Momentum

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$A$ shell of mass $5 M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M, 2 M$, and $2 M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes $2 v$ and $v$ respectively. The third fragment will

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A
move with a velocity $v$ in a direction perpendicular to the other two
B
move with a velocity $2 v$ in the direction of velocity of the first fragment
C
be at rest
D
move with velocity $v$ in the direction of velocity of the second fragment

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Newton's Laws of Motion and Friction

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Conservation of Linear Momentum

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$A$ shell of mass $5 M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M, 2 M$, and $2 M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes $2 v$ and $v$ respectively. The third fragment will

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