A ball with a mass of #4 kg # and velocity of #6 m/s# collides with a second ball with a mass of #7 kg# and velocity of #- 1 m/s#. If #25%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-22T11:07:20

The final velocities are #=-2.91ms^-1# and #=1.30ms^-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,

#4xx6+7xx(-1)=4v_1+7v_2#

#4v_1+7v_2=17#

#v_2=(17-4v_1)/(7)#........................#(1)#

and

#0.75(1/2xx4xx6^2+1/2xx7xx(-1)^2)=1/2xx4xxv_1^2+1/2xx7xxv_2^2#

#4v_1^2+7v_2^2=151#...................#(2)#

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

#4v_1^2+7*((17-4v_1)/(7))^2=151#

#28v_1^2+16v_1^2-136v_1+289-1057=0#

#44v_1^2-136v_1-768=0#

#11v_1^2-34v_1-192=0#

Solving this quadratic equation in #v_1#

#v_1=(34+-sqrt(34^2-4xx11xx-192))/(22)#

#v_1=(34+-sqrt(9604))/(22)#

#v_1=(34+-98)/(22)#

#v_1=6ms^-1# or #v_1=-2.91ms^-1#

#v_2=-1ms^-1# or #v_2=1.30ms^-1#