A ball with a mass of 5 kg5kg is rolling at 9 m/s9ms and elastically collides with a resting ball with a mass of 2 kg2kg. What are the post-collision velocities of the balls?

1 Answer
Apr 8, 2017

The velocity of the first ball is =3.86ms^-1=3.86ms1
The velocity of the second ball is =12.86ms^-1=12.86ms1

Explanation:

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

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

m_1=5kgm1=5kg

m_2=2kgm2=2kg

u_1=9ms^-1u1=9ms1

u_2=0ms^-1u2=0ms1

Therefore,

v_1=3/7*9+4/7*(0)=27/7=3.86ms^-1v1=379+47(0)=277=3.86ms1

v_2=10/7*9-3/7*(0)=90/7=12.86ms^-1v2=107937(0)=907=12.86ms1

Verificaition

m_1u_1+m_2u_2=5*9+2*0=45m1u1+m2u2=59+20=45

m_1v_1+m_2v_2=5*3.86+2*12.86=45m1v1+m2v2=53.86+212.86=45