A ball with a mass of #4 kg # and velocity of #3 m/s# collides with a second ball with a mass of #2 kg# and velocity of #- 1 m/s#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

1470 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-26T15:05:44

The final velocities are #=0.85ms^-1# and #=3.3ms^-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,

#4xx3+2xx(-1)=4v_1+2v_2#

#4v_1+2v_2=10#

#v_2=(5-2v_1)#........................#(1)#

and

#0.25(1/2xx4xx3^2+1/2xx2xx(-1)^2)=1/2xx4xxv_1^2+1/2xx2xxv_2^2#

#4v_1^2+2v_2^2=9.5#

#2v_1^2+v_2^2=4.75#...................#(2)#

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

#2v_1^2+((5-2v_1))^2=4.75#

#2v_1^2+4v_1^2-20v_1+25-4.75=0#

#6v_1^2-20v_1-20.25=0#

Solving this quadratic equation in #v_1#

#v_1=(20+-sqrt(20^2-4xx6xx-20.25))/(12)#

#v_1=(40+-sqrt(886))/(12)#

#v_1=(40+-29.8)/(12)#

#v_1=5.81ms^-1# or #v_1=0.85ms^-1#

#v_2=-6.62ms^-1# or #v_2=3.3ms^-1#