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?

1 Answer
Narad T.
Jun 26, 2017

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

Explanation:

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

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