Body A of mass 4m moving with speed u collides with another body B of mass 2m at test. The collision is head on and elastic in nature. After the collision, the fraction of energy lost by the colliding body A is  [NEET 2019]

detailed solution

Correct option is A

The given situation of collision can be drawn asApplying law of conservation of linear momentum, Initial momentum of system = Final momentum of system⇒        (4m)u+(2m)u'=(4m)v1+(2m)v2                 4mu + (2m)×0 = 4mv1 + 2mv2or                              2u=2v1+v2                                                   …(i)The kinetic energy of A before collision,                                (KE)A=12(4m)u2=2mu2Kinetic energy of B before collision,                            (KE)B = 0The kinetic energy of A after collision,            KE'A=12(4m)v12=2mv12Kinetic energy of B after collision,               (KE')B=12(2m)v22=mv22As, initial kinetic energy of the system = final kinetic energy of the system⇒               (KE)A+(KE)B=KE'A+KE'B                                 2mu2 + 0 = 2mv12 + mv22                                        2mu2=2mv12 + mv22or                                        2u2=2v12 + v22                                   …(ii)Solving Eqs. (j) and (ii), we get                             v1=13u  and v2=43uor the final velocity of A can be directly calculated by using the formula,                  v1=m1-m2m1+m2u1+2m2u2m1+m2                        =4m-2m4m+2mu+2(2m)×0(4m+2m)                  ∵ u2=u'=0                        =2m6mu=13u∴  Net decrease in kinetic energy of A,            ΔKE=(KE)A-KE'A=2mu2-2mv12                     = 2m (u2-v12)Substituting the value of v1, we get                   ΔKE=2mu2-u29=16mu29∴  The fractional decrease in kinetic energy,             ΔKE(KE)A=16mu29×12mu2=89