Question #6d21b

1 Answer
Jan 31, 2018

Answer:

The velocities after collision are #=(v_1(m_1-m_2)+2v_2m_2)/(m_2+m_1)# and #(v_2(m_2-m_1)+2m_1v_1)/(m_1+m_2)#

Explanation:

In an elastic collision, there is conservation of momentum and conservation of kinetic energy

By conservation of momentum,

#m_1v_1+m_2v_2=m_1w_1+m_2w_2#...................#(1)#

By the conservation of kinetic energy,

#1/2m_1v_1^2+1/2m_2v_2^2=1/2m_1w_1^2+1/2m_2w_2^2#

#m_1v_1^2+m_2v_2^2=m_1w_1^2+m_2w_2^2#...................#(2)#

Solving for #w_1# and #w_2# in equations #(1)# and #(2)#

From #(1)#

#m_1(v_1-w_1)=m_2(w_2-v_2)#......................#(3)#

From #(2)#

#m_1(v_1^2-w_1^2)=m_2(w_2^2-v_2^2)#

#m_1(v_1+w_1)(v_1-w_1)=m_2(w_2-v_2)(w_2+v_2)#.....#(4)#

From #(3)# and #(4)#

#v_1+w_1=v_2+w_2#

#w_1=v_2-v_1+w_2#

Plugging this value in #(1)#

#m_1v_1+m_2v_2=m_1(v_2-v_1+w_2)+m_2w_2#

Solving for #w_2#

#m_1v_1+m_2v_2=m_1v_2-m_1v_1+m_1w_2+m_2w_2#

#w_2(m_1+m_2)=v_2(m_2-m_1)+2m_1v_1#

#w_2=(v_2(m_2-m_1)+2m_1v_1)/(m_1+m_2)#

Therefore,

#w_1=v_2-v_1+(v_2(m_2-m_1)+2m_1v_1)/(m_1+m_2)#

#=((v_2-v_1)(m_1+m_2)+(v_2(m_2-m_1)+2m_1v_1))/(m_1+m_2)#

#=(v_2m_1+v_2m_2-v_1m_1-v_1m_2+v_2m_2-v_2m_1+2m_1v_1)/(m_1+m_2)#

#=(v_1(m_1-m_2)+2v_2m_2)/(m_2+m_1)#