A particle of mass m1 makes an elastic head on collision with a stationary particle of mass m2. the fraction of kinectic energy of m1 carried by m2?

1 Answer
Mar 2, 2018

Let initial velocity of particle m_1 be u_1, final velocity of particle m_1 be v_1 and that of particle m_2 be v_2.

Since collision is an elastic head on, from Law of Conservation of Momentum we have

m_1u_1 = m_1 v_1 + m_2 v_2
=>m_2 v_2=m_1u_1 - m_1 v_1 .........(1)

From the Law of Conservation of kinetic energy we have

1/2 m_1 u_1^2 = 1/2 m_1 v_1^2 + 1/2 m_2 v_2^2 .......(2)

Rewriting (1) as

m_2v_2=m_1(u_1 - v_1 ) .......(3)

Rewriting (2) as

m_2 v_2^2= m_1 (u_1^2-v_1^2) .......(4)

Dividing (4) with (3)

v_2=u_1+v_1 .......(2)

Eliminating v_2 from (1) and (5) we get

(m_1u_1 - m_1 v_1)/m_2=u_1+v_1
=>(m_1u_1 - m_1 v_1)=m_2u_1+m_2v_1
=>m_1u_1 -m_2u_1 =m_1 v_1+m_2v_1
=>(m_1 -m_2)/(m_1+m_2)u_1 =v_1 .........(6)

Fraction of KE of m_1 carried by m_2

(KE_i-KE_f)/(KE_i)=1-(KE_f)/(KE_i)=1-(1/2m_1v_1^2)/(1/2m_1u_1^2)=1-(v_1/u_1)^2

Using (6) in above we get

Fraction of KE of m_1 carried by m_2 is

1-[(m_1 -m_2)/(m_1+m_2)]^2=(4m_1m_2)/((m_1+m_2)^2)