A ball with a mass of #6 kg # and velocity of #8 m/s# collides with a second ball with a mass of #5 kg# and velocity of #- 2 m/s#. If #10%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-03-02T10:18:28

The final velocities are #=-0.61ms^-1# and #=8.33ms^-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,

#6xx8+5xx(-2)=6v_1+5v_2#

#6v_1+5v_2=38#

#5v_2=38-6v_1#

#v_2=((38-6v_1))/5#........................#(1)#

and

#0.9(1/2xx6xx8^2+1/2xx5xx(-2)^2)=1/2xx6xxv_1^2+1/2xx5xxv_2^2#

#6v_1^2+5v_2^2=363.6#...................#(2)#

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

#6v_1^2+5(((38-6v_1))/5)^2=363.6#

#30v_1^2+36v_1^2-576v_1+1444-1818=0#

#66v_1^2-576v_1-374=0#

#33v_1^2-288v_1-187=0#

Solving this quadratic equation in #v_1#

#v_1=(288+-sqrt(288^2+4xx33xx(187)))/(2*33)#

#v_1=(288+-sqrt(107628))/(66)#

#v_1=(288+-328.1)/(66)#

#v_1=9.33ms^-1# or #v_1=-0.61ms^-1#

#v_2=-3.60ms^-1# or #v_2=8.33ms^-1#