A ball with a mass of # 3 kg# is rolling at #2 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 · 2017-06-26T10:13:42

The velocity of the first ball is #=-2/7ms^-1# and the second ball is #=12/7ms^-1#

Since the collision is elastic, there is conservation of linear momentum and conservation of kinetic energy.

#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+1/2mv_2^2#

#m_1=3kg#

#m_2=4kg#

#u_1=2ms^-1#

#u_2=0 ms^-1#

So,

#3*2+4*0=3*v_1+4*v_2#

#3v_1+4v_2=6#...............................#(1)#

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

#3v_1^2+4v_2^2=12#........................#(2)#

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

#v_2=(6-3v_1)/4#

#3v_1^2+4*((6-3v_1)/4)^2=12#

#12v_1^2+36-36v_1+9v_1^2=48#

#21v_1^2-36v_1-12=0#

#7v_1^2-12v_1-4=0#

#v_1=(12+-sqrt(12^2+4*4*7))/(14)#

#=(12+-16)/14#

#v_1=2ms^-1# or #v_1=-2/7=-0.28ms^-1#

We keep the second result #v_1=0.28ms^-1#

#v_2=(6+3*0.28)/4=12/7=1.71#

Verification

#4*12/7-3*2/7=6#