A ball with a mass of #2 kg # and velocity of #4 m/s# collides with a second ball with a mass of #5 kg# and velocity of #- 6 m/s#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-20T10:51:19

The final velocities are #=-10.3ms^-1# and #=0.1ms^-1#

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

The kinetic energy is

#k(1/2m_1u_1^2+1/2m_2u_2^2)=1/2m_1v_1^2+1/2m_2v_2^2#

Therefore,

#2xx4+5xx(-6)=2v_1+5v_2#

#2v_1+5v_2=-22#

#v_2=-((2v_1+22))/5#........................#(1)#

and

#0.8(1/2xx2xx4^2+1/2xx5xx(-6)^2)=1/2xx2xxv_1^2+1/2xx5xxv_2^2#

#2v_1^2+5v_2^2=212#...................#(2)#

Solving for #v_1# and #v_2# in equation s #(1)# and #(2)#

#2v_1^2+5*(-((2v_1+22))/5)^2=212#

#10v_1^2+4v_1^2+88v_1+484-1060=0#

#14v_1^2+88v_1-576=0#

#7v_1^2+44v_1-288=0#

Solving this quadratic equation in #v_1#

#v_1=(-44+-sqrt(44^2-4xx7xx-288))/(14)#

#v_1=(-44+-sqrt(10000))/(14)#

#v_1=(-44+-100)/(14)#

#v_1=4ms^-1# or #v_1=-10.3ms^-1#

#v_2=-6ms^-1# or #v_2=0.1ms^-1#