A ball with a mass of #4 kg # and velocity of #1 m/s# collides with a second ball with a mass of #9 kg# and velocity of #- 4 m/s#. If #40%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-01-27T10:46:19

The final velocities are #=-3.78ms^-1# and #=-1.88ms^-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,

#4xx1+9xx(-4)=4v_1+9v_2#

#4v_1+9v_2=-32#

#v_2=(-(32+4v_1))/9#........................#(1)#

and

#0.60(1/2xx4xx1^2+1/2xx9xx(-4)^2)=1/2xx4xxv_1^2+1/2xx9xxv_2^2#

#4v_1^2+9v_2^2=88.8#...................#(2)#

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

#4v_1^2+9((-(32+4v_1))/9)^2=88.8#

#36v_1^2+16v_1^2+256v_1+1024-799.2=0#

#52v_1^2+256v_1+224.8=0#

#13v_1^2+64v_1+56.2=0#

Solving this quadratic equation in #v_1#

#v_1=(-64+-sqrt(64^2-4xx13xx56.2))/(26)#

#v_1=(-64+-sqrt(1173.6))/(26)#

#v_1=(-64+-34.3)/(26)#

#v_1=-3.78ms^-1# or #v_1=-1.14ms^-1#

#v_2=-1.88ms^-1# or #v_2=-3.05ms^-1#