A ball with a mass of #6 kg # and velocity of #5 m/s# collides with a second ball with a mass of #3 kg# and velocity of #- 2 m/s#. If #50%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

1141 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-22T13:14:43

The final velocities are #=-0.975ms^-1# and #=9.95ms^-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,

#6xx5+3xx(-2)=6v_1+3v_2#

#6v_1+3v_2=24#

#v_2=(8-2v_1)#........................#(1)#

and

#0.5(1/2xx6xx5^2+1/2xx3xx(-2)^2)=1/2xx6xxv_1^2+1/2xx3xxv_2^2#

#6v_1^2+3v_2^2=81#

#2v_1^2+v_2^2=27#...................#(2)#

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

#2v_1^2+((8-2v_1))^2=27#

#2v_1^2+4v_1^2-32v_1+64-27=0#

#6v_1^2-32v_1-37=0#

Solving this quadratic equation in #v_1#

#v_1=(32+-sqrt(32^2-4xx6xx-37))/(12)#

#v_1=(32+-sqrt(1912))/(12)#

#v_1=(32+-43.7)/(12)#

#v_1=6.31ms^-1# or #v_1=-0.975ms^-1#

#v_2=-4.62ms^-1# or #v_2=9.95ms^-1#