A stationary particle breaks into two parts of masses $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$ which move with velocities $\mathrm{v}_{\mathrm{A}}$ and $\mathrm{v}_{\mathrm{B}}$ respectively. The ratio of their kinetic energies ($\mathrm{K}_{\mathrm{B}}:\mathrm{K}_{\mathrm{A}}$) is :

(A) $\mathrm{v}_{\mathrm{B}}:\mathrm{v}_{\mathrm{A}}$

(C) $\mathrm{m}_{\mathrm{B}}\mathrm{v}_{\mathrm{B}}:\mathrm{m}_{\mathrm{A}}\mathrm{v}_{\mathrm{A}}$

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Conservation of Momentum

A stationary particle breaks into two parts of masses

m_{A}

and

m_{B}

which move with velocities

v_{A}

and

v_{B}

respectively. The ratio of their kinetic energies (

K_{B} : K_{A}

) is :

(A) $v_{B} : v_{A}$ ✓ Correct
(B) $m_{B} : m_{A}$
(C) $m_{B} v_{B} : m_{A} v_{A}$
(D) $1 : 1$

Answer: (A)

Solution

Momentum conservation from rest:

m_{A} v_{A} = m_{B} v_{B}

. Then

\frac{K _{B}}{K _{A}} = \frac{\frac{1}{2} m _{B} v _{B}^{2}}{\frac{1}{2} m _{A} v _{A}^{2}} = \frac{( m _{B} v _{B} ) v _{B}}{( m _{A} v _{A} ) v _{A}} = \frac{v _{B}}{v _{A}}

K_{B} : K_{A} = v_{B} : v_{A}

. Option (A).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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