A ball with a mass of #5 kg # and velocity of #6 m/s# collides with a second ball with a mass of #3 kg# and velocity of #- 7 m/s#. If #40%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-29T16:13:50

The final velocities are #=-3.18ms^-1# and #=8.27ms^-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,

#5xx6+3xx(-7)=5v_1+3v_2#

#5v_1+3v_2=9#

#v_2=((9-5v_1))/3#........................#(1)#

and

#0.60(1/2xx5xx6^2+1/2xx3xx(-7)^2)=1/2xx5xxv_1^2+1/2xx3xxv_2^2#

#5v_1^2+3v_2^2=196.2#...................#(2)#

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

#5v_1^2+3(((9-5v_1))/3)^2=196.2#

#15v_1^2+25v_1^2-90v_1+81-196.2=0#

#40v_1^2-90v_1-115.2=0#

#20v_1^2-45v_1-57.6=0#

Solving this quadratic equation in #v_1#

#v_1=(-45+-sqrt(45^2+4xx20xx57.6))/(40)#

#v_1=(-45+-sqrt(6633))/(40)#

#v_1=(-45+-81.44)/(40)#

#v_1=-3.16ms^-1# or #v_1=0.91ms^-1#

#v_2=8.27ms^-1# or #v_2=-1.48ms^-1#