A ball with a mass of # 3 kg# is rolling at #8 m/s# and elastically collides with a resting ball with a mass of #4 kg#. What are the post-collision velocities of the balls?

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Answer 1 · 2018-01-05T09:37:18

The post collision velocities are #=-1.14ms^-1# and #=6.86ms^-1#

In an elastic collision, there is conservation of momentum and conservation of kinetic energies.

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

#1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2#

Here,

The masses are

#m_1=3kg#

and #m_2=4kg#

The initial velocities are

#u_1=8ms^-1#

and #u_2=0ms^-1#

#3*8+4*0=3*v_1+4*v_2#, #<=>#, #3v_1+4v_2=24#........#(1)#

#1/2*3*8^2+1/2*4*0^2=1/2*3*v_1^2+1/2*4*v_2^2#

#3v_1^2+4v_2^2=192#.............................#(2)#

Solving for #v_1# and #v_2# in equations #(1)# and #(2)#

From #(1)#, #v_2=(24-3v_1)/4#

Substituting is #(2)#

#3v_1^2+4*((24-3v_1)/4)^2=192#

#12v_1^2+(24-3v_1)^2=768#

#12v_1^2+576-144v_1+9v_1^2=768#

#21v_1^2-144v-192=0#

#7v_1^2-48v-64=0#

#v_1=(48+-sqrt((-48)^2-4*7*(-64)))/(2*7)#

#=(48+-sqrt(4096))/(14)#

#=(48+-64)/(14)#

So,

#v_1=((48+64)/14)=8ms^-1#, #=>#, #v_2=0ms^-1#

This is the initial conditions.

or

#v_2=((48-64)/14)=-1.14ms^-1#, #=>#, #v_2=6.86ms^-1#