A ball with a mass of 2 kg2kg is rolling at 4 m/s4ms and elastically collides with a resting ball with a mass of 4 kg4kg. What are the post-collision velocities of the balls?

1 Answer
May 9, 2017

The velocity of the first ball is =-1.33ms^-1=1.33ms1
The velocity of the second ball is =2.67ms^-1=2.67ms1

Explanation:

In an elastic collision, we have conservation of momentum and conservation of kinetic energy.

The velocities before the collision are u_1u1 and u_2u2.

The velocities after the collision are v_1v1 and v_2v2.

m_1u_1+m_2u_2=m_1v_1+m_2v_2m1u1+m2u2=m1v1+m2v2

and

1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^212m1u21+12m2u22=12m1v21+12m2v22

Solving the above 2 equations for v_1v1 and v_2v2, we get

v_1=(m_1-m_2)/(m_1+m_2)*u_1+(2m_2)/(m_1+m_2)*u_2v1=m1m2m1+m2u1+2m2m1+m2u2

and

v_2=(2m_1)/(m_1+m_2)*u_1+(m_2-m_1)/(m_1+m_2)*u_2v2=2m1m1+m2u1+m2m1m1+m2u2

Taking the direction as positive rarr^++

m_1=2kgm1=2kg

m_2=4kgm2=4kg

u_1=4ms^-1u1=4ms1

u_2=0ms^-1u2=0ms1

Therefore,

v_1=-2/6*4+8/6*(0)=-4/3=-1.33ms^-1v1=264+86(0)=43=1.33ms1

v_2=4/6*4-2/6*(0)=8/3=2.67ms^-1v2=46426(0)=83=2.67ms1

Verificaition

m_1u_1+m_2u_2=2*4+4*0=8m1u1+m2u2=24+40=8

m_1v_1+m_2v_2=-2*4/3+4*8/3=8m1v1+m2v2=243+483=8