A ball with a mass of 5 kg5kg is rolling at 12 m/s12ms and elastically collides with a resting ball with a mass of 4 kg4kg. What are the post-collision velocities of the balls?

1 Answer
Nov 23, 2017

The velocities are 4/3ms^-143ms1 and 40/3ms^-1403ms1

Explanation:

Here it's an elastic collision with no loss of kinetic energy.

We have the conservation of momentum.

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

m_1=5 kgm1=5kg

u_1=12 ms^(-1)u1=12ms1

m_2=4 kgm2=4kg

u_2=0u2=0

v_1=?v1=?

v_2=?v2=?

5*12+4*0=5*v_1+4*v_2512+40=5v1+4v2

5v_1+4v_2=605v1+4v2=60............(1)(1)

There is also conservation of kinetic energies

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

5*12^2+4*0=5*v_1^2+4*v_2^25122+40=5v21+4v22

5v_1^2+4v_2^2=7205v21+4v22=720.........(2)(2)

From (1)(1), we get

v_1=(60-4v_2)/5v1=604v25

Plugging this value in (2)(2)

5*((60-4v_2)/5)^2+4v_2^2=7205(604v25)2+4v22=720

((60-4v_2))^2/5+4v_2^2=720((604v2))25+4v22=720

16(15-v_2)^2+20v_2^2=720*5=360016(15v2)2+20v22=7205=3600

4(15-v_2)^2+5v_2^2=9004(15v2)2+5v22=900

4(225-30v_2+v^2)+5v_2^2=9004(22530v2+v2)+5v22=900

900-120v_2+4v_2^2+5v_2^2=900900120v2+4v22+5v22=900

9v_2^2-120v_2=09v22120v2=0

v_2(9v_2-120)=0v2(9v2120)=0

v_2=0v2=0 or v_2=120/9=40/3=13.3ms^-1v2=1209=403=13.3ms1

v_1=(60-4*40/3)/5=4/3ms^-1v1=6044035=43ms1