A ball with a mass of #4 kg # and velocity of #3 m/s# collides with a second ball with a mass of #5 kg# and velocity of #- 1 m/s#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

1884 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-02T10:15:50

The final velocities are #=0.38ms^-1# and #=1.1ms^-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+5xx(-1)=4v_1+5v_2#

#4v_1+5v_2=7#

#5v_2=7-4v_1#

#v_2=((7-4v_1))/5#........................#(1)#

and

#0.8(1/2xx4xx3^2+1/2xx5xx(-1)^2)=1/2xx4xxv_1^2+1/2xx5xxv_2^2#

#4v_1^2+5v_2^2=32.8#...................#(2)#

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

#4v_1^2+5(((7-4v_1))/5)^2=32.8#

#20v_1^2+16v_1^2-56v_1+49-32.8=0#

#36v_1^2-56v_1+16.2=0#

#18v_1^2-28v_1+8.1=0#

Solving this quadratic equation in #v_1#

#v_1=(28+-sqrt(28^2-4xx18xx(8.1)))/(2*18)#

#v_1=(28+-sqrt(200.8))/(36)#

#v_1=(28+-14.2)/(36)#

#v_1=1.17ms^-1# or #v_1=0.38ms^-1#

#v_2=0.46ms^-1# or #v_2=1.1ms^-1#