Question

a body in velocity having mass m collides head on elastically with another body at rest of mass M. for maximum transfer of speed. also explain the reason: 1. m=M 2.m>>M 3. m<

Answer 1

If two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i .

Answer 2

Answer:

Using momentum conservation and equation for coefficient of restitution in case of elastic collision, we get,

v1=m1+m2m1−m2uandv2=m1+m22m1u

where u is the velocity of m1 and m2 is at rest.

Kinetic energy of m1=1/2×m1v12

Therefore, fraction of energy retained =(m1+m2m1−m2)2

Kinetic energy of m2=1/2×m2v22

Therefore, fraction of energy transferred =(m1+m2)24m1m2.

Also when both the masses are equal, the velocity gets interchanged. Therefore, 100% energy is transferred. No energy is lost in elastic collisions.