# 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,

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

${V}_{1}=\left(\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\right){U}_{1}+\left(\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}\right){U}_{2}\phantom{\rule{0ex}{0ex}}and\phantom{\rule{0ex}{0ex}}{V}_{2}=\left(\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}\right){U}_{2}+\left(\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}\right){U}_{1}$

Where,
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

Therefore,

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

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

Therefore, the percentage of the kinetic energy transferred is,