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
A08
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)#

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