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

Question

1546 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-02T17:02:47

This is not possible.

#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,

#1xx3+3xx(-4)=1v_1+3v_2#

#v_1+3v_2=-9#

#3v_2=-9-v_1#

#v_2=(-(9+v_1))/3#........................#(1)#

and

#0.25(1/2xx1xx3^2+1/2xx3xx(-4)^2)=1/2xx1xxv_1^2+1/2xx3xxv_2^2#

#v_1^2+3v_2^2=14.25#...................#(2)#

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

#v_1^2+3((-((9+v_1))/3)^2=14.25#

#3v_1^2+v_1^2+18v_1+81-42.75=0#

#4v_1^2+18v_1+38.25=0#

Solving this quadratic equation in #v_1#

#v_1=(-18+-sqrt(18^2-4xx4xx(38.25)))/(2*18)#

The discriminant is negative, there is no solution.