Work Energy Power

JEE Main level, single correct, medium

52 of 183

Question 52 of 183

A stationary particle breaks into two parts of masses mAm_A and mBm_B which move with velocities vAv_A and vBv_B respectively. The ratio of their kinetic energies (KB:KA)\left(K_B: K_A\right) is :

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Answer: (A) vB:vAv_B: v_A

Initial momentum is zero.
 Hence ∣PA∣=∣PB∣⇒mAvB=mBvB(KE)A(KE)B=12 mAvA212 mBvB2=vAvB(KE)B(KE)A=vBvA\begin{aligned} & \text { Hence }\left|\mathrm{P}_{\mathrm{A}}\right|=\left|\mathrm{P}_{\mathrm{B}}\right| \\ & \Rightarrow \mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}} \\ & \frac{(\mathrm{KE})_{\mathrm{A}}}{(\mathrm{KE})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^2}{\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^2}=\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}} \\ & \frac{(\mathrm{KE})_{\mathrm{B}}}{(\mathrm{KE})_{\mathrm{A}}}=\frac{\mathrm{v}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{A}}}\end{aligned}

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