A ball with a mass of #6 kg # and velocity of #1 m/s# collides with a second ball with a mass of #4 kg# and velocity of #- 7 m/s#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-30T13:21:25

The final velocities are #=-2.57ms^-1# and #=1.65ms^-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,

#6xx1+4xx(-7)=6v_1+4v_2#

#6v_1+4v_2=-22#

#3v_1+2v_2=-11#

#v_2=-((11+3v_1))/2#........................#(1)#

and

#0.25(1/2xx6xx1^2+1/2xx4xx(-7)^2)=1/2xx6xxv_1^2+1/2xx4xxv_2^2#

#6v_1^2+4v_2^2=50.5#...................#(2)#

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

#6v_1^2+4((-(11+3v_1))/2)^2=50.5#

#6v_1^2+9v_1^2+66v_1+121-50.5=0#

#15v_1^2+66v_1+70.5=0#

#5v_1^2+22v_1+23.5=0#

Solving this quadratic equation in #v_1#

#v_1=(-22+-sqrt(22^2-4xx5xx23.5))/(10)#

#v_1=(-22+-sqrt(14))/(10)#

#v_1=(-22+-3.74)/(10)#

#v_1=-2.57ms^-1# or #v_1=1.83ms^-1#

#v_2=1.65ms^-1# or #v_2=-8.25ms^-1#