A ball with a mass of #6 kg # and velocity of #5 m/s# collides with a second ball with a mass of #1 kg# and velocity of #- 7 m/s#. If #50%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

1286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-23T10:35:34

The velocities of the balls are #=4.04ms^-1# and #=1.24ms^-1#

Momentum is conserved

#m_1u_1+m_2u_2=m_1*v_1+m_2v_2#

#u_1=5ms^-1#

#u_2=-7ms^-1#

#m_1=6kg#

#m_2=1kg#

#6*5-1*7=6v_1+1*v_2#

#6v_1+v_2=23#--------------#(1)#

#KE# before is #=1/2m_1u_1^2+1/2m_2u_2^2#

#=1/2*6*5^2+1/2*1*(-7)^2=75+49/2=99.5#

#KE# after is #=1/2m_1v_1^2+1/2m_2v_2^2#

#=1/2*6v_1^2+1/2*1v_2^2=99.5/2#

So,

#6v_1^2+v_2^2=99.5#------------------#(2)#

We have 2 equations with 2 unknowns

From #(1)#

#v_2=23-6v_1#

Replacing in #(2)#

#6v_1^2+(23-6v_1)^2=99.5#

#6v_1+529-276v_1+36v^2=99.5#

#42v_1^2-276v_1+429.5=0#

#v_1=(276+-sqrt(276^2-4*42*429.5))/84#

#v_1=(276+-sqrt4020)/84#

#v_1=2.6# or #v_1=4.04#

#v_2=7.4# or #v_2=-1.24#

The velocities of the balls are #=4.04ms^-1# and #=1.24ms^-1#