A ball with a mass of #3 kg # and velocity of #5 m/s# collides with a second ball with a mass of #7 kg# and velocity of #- 2 m/s#. If #25%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-02-01T13:16:21

The final velocities are #=-2.78ms^-1# and #=1.33ms^-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,

#3xx5+7xx(-2)=3v_1+7v_2#

#3v_1+7v_2=1#

#7v_2=1-3v_1#

#v_2=((1-3v_1))/7#........................#(1)#

and

#0.75(1/2xx3xx5^2+1/2xx7xx(-2)^2)=1/2xx3xxv_1^2+1/2xx7xxv_2^2#

#3v_1^2+7v_2^2=77.25#...................#(2)#

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

#3v_1^2+7(((1-3v_1))/7)^2=77.25#

#21v_1^2+63v_1^2-42v_1+7-540.75=0#

#84v_1^2-42v_1-533.75=0#

Solving this quadratic equation in #v_1#

#v_1=(-42+-sqrt(42^2-4xx84xx(-533.75)))/(2*84)#

#v_1=(-42+-sqrt(181104))/(168)#

#v_1=(-42+-425.56)/(168)#

#v_1=-2.78ms^-1# or #v_1=2.28ms^-1#

#v_2=1.33ms^-1# or #v_2=-0.83ms^-1#