A ball with a mass of #3 kg # and velocity of #1 m/s# collides with a second ball with a mass of #4 kg# and velocity of #- 7 m/s#. If #50%# of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-02-05T10:38:16

The final velocities are #=4.97ms^-1# and #=2.52ms^-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,

#3xx1+4xx(-7)=3v_1+4v_2#

#3v_1+4v_2=-25#

#4v_2=-25-3v_1#

#v_2=(-(25+3v_1))/4#........................#(1)#

and

#0.50(1/2xx3xx1^2+1/2xx4xx(-7)^2)=1/2xx3xxv_1^2+1/2xx4xxv_2^2#

#3v_1^2+4v_2^2=99.5#...................#(2)#

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

#3v_1^2+4((-((25+3v_1))/4)^2)=99.5#

#12v_1^2+9v_1^2+150v_1+625-398=0#

#21v_1^2+150v_1+227=0#

Solving this quadratic equation in #v_1#

#v_1=(-150+-sqrt(150^2-4xx21xx(227)))/(2*21)#

#v_1=(-150+-sqrt(3432))/(42)#

#v_1=(-150+-58.6)/(42)#

#v_1=-2.18ms^-1# or #v_1=-4.97ms^-1#

#v_2=-4.62ms^-1# or #v_2=2.52ms^-1#