JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Conservation of Momentum

An artillery piece of mass

M_{1}

fires a shell of mass

M_{2}

horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:

(A) $M_{1} / (M_{1} + M_{2})$
(B) $\frac{M _{2}}{M _{1}}$ ✓ Correct
(C) $M_{2} / (M_{1} + M_{2})$
(D) $\frac{M _{1}}{M _{2}}$

Answer: (B)

Solution

By conservation of momentum, the artillery (

M_{1}

) and shell (

M_{2}

) acquire equal and opposite momenta:

p_{1} = p_{2} = p

.

Since

K E = \frac{p ^{2}}{2 M}

for the same

p

\frac{K E _{1}}{K E _{2}} = \frac{p ^{2} /2 M _{1}}{p ^{2} /2 M _{2}} = \frac{M _{2}}{M _{1}}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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