A ball with a mass of #5 kg # and velocity of #2 m/s# collides with a second ball with a mass of #3 kg# and velocity of #- 5 m/s#. If #10%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-21T08:22:57

The final velocities are #0.81ms^-1# and #-3.02ms^-1# or #-2.68ms^-1# and #2.8ms^-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,

#5xx2+3xx(-5)=5v_1+3v_2#

#5v_1+3v_2=-5#

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

and

#0.9(1/2xx5xx2^2+1/2xx3xx(-5)^2)=1/2xx5xxv_1^2+1/2xx3xxv_2^2#

#5v_1^2+3v_2^2=85.5#...................#(2)#

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

#5v_1^2+3*(-(5(v_1+1))/3)^2=85.5#

#15v_1^2+25v_1^2+50v_1+25-256.5=0#

#40v_1^2+75v_1-231.5=0#

#8v_1^2+15v_1-46.3=0#

Solving this quadratic equation in #v_1#

#v_1=(-15+-sqrt(15^2-4xx3xx-46.3))/(16)#

#v_1=(-15+-sqrt(780.6))/(16)#

#v_1=(-15+-27.9)/(16)#

#v_1=0.81ms^-1# or #v_1=-2.68ms^-1#

#v_2=-3.02ms^-1# or #v_2=2.8ms^-1#