a particle of mass m moving with a certain velocity collides elastically head on with a particle of mass 4m at rest. the percentage of K.E transferred is? (ans : 64%)

Dear Student,

Please find below the solution to the asked query:

By using conservation of momentum and conservation of energy, the equations for the velocities of the spheres after collision are,


m1 = mass of the first particle,
m2 = mass of the second particle,
U1 = speed of first particle before collision,
U2 = speed of second particle before collision,
V1 = speed of first particle after collision
V2 = speed of second particle after collision


The kinetic energy transferred is equal to the kinetic energy of the second particle after collision. Therefore,

V2=m2-m1m1+m2U2+2m1m1+m2U1  V2=4m-mm+4m0+2mm+4mU  V2=2U5

So, the kinetic energy of the second particle after the collision is,

K.E = 124mV22 = 124m2U52  K.E= 162512mU2
Therefore, the percentage of the kinetic energy transferred is,

K.E % = K.E2K.Ei,tot×100 = 162512mU212mU2×100 = 1625×100 K.E % = 64 %


Hope this information will clear your doubts about the topic.

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