A particle of mass m moving with speed u collides perfectly inelastically with another particle of mass 2m at rest .find loss of kinetic energy of system in the collisions

Dear Student,
 

Let the first particle be of mass mA=m and initial speed vA=u.Let the first particle be of mass mB=2m and initial speed vB=0.It is given that the first particle collides perfectly inelastically with the second particle.They stick together.Let their combined mass be msys=(m+2m)=3m; and their final velocity be vsys.Now, from the law of conservation of momentum,initial momentum of the system = final momentum of the system.mAvA+mBvB=msysvsys(m·u) + (2m·0)=3m·vsysmu=3m·vsysvsys=m3m·uvsys=u3
The initial kinetic energy = 12mAvA2 + 12mBvB2Initial K.E. = 12m·u2 + 12(2m)·02 = 12mu2 The final kinetic energy = 12msysvsys2Final K.E. =12(3m)·u32=12·3mu29=16mu2Fractional loss in K.E.= Initial K.E. - Final K.E.Initial K.E.×100 =12mu2 -16mu212mu2 ×100=13mu2 12mu2 ×100=23×10067%
 
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