A moving particle collides with a stationary particle of mass 1n times the mass of moving particle, the fraction of its kinetic energy transferred to the stationary particle is

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Accepted Answer

Correct answer: B. (c) By law of conservation of momentum P_i=P_f ⇒ mV+O=mV_1+ mnV_2 ⇒V=V_1+ V_2n As nothing is cleared about type of collision, consider collision as elastic. So, e=1 ⇒ V_2-V_1 V=1 ⇒V_2-V_1=V So , V_2-(V- V_2n)=VV_2(1+ 1n)=2VV_2= 2 Vnn+1 So, K= 12 mn 4 V^2n^2 (n+1 )^2. So, KK_f= 12 mn ( 4 V^2n^2 (n+1 )^2 ) 12 mV^2= 4n1+n^2

$A moving particle collides with a stationary particle of mass times the mass of moving particle, the fraction of its kinetic energy transferred to the stationary particle is$

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