Three balls roll on a frictionless surface and collide in two separate collisions. Ball 1 (mass m1=1.0 kg and velocity v1i=10 m/s) collides first with Ball 2 (mass m2=2 kg and velocity v2i=0 m/s) in an inelastic collision. Ball 2 then collides with Ball 3 (mass m3=0.9 kg and velocity v3i=5 m/s) in a perfectly inelastic collision. If Ball 1 moves with a velocity v1f=2.5 m/s to the left after its collision, what is the final velocity (v3f) of the combined balls (2 and 3) after their collision in m/s? Assume all three balls move only in the x direction. If your answer is negative (moving to the left), include the negative in your entry. Round your answer to 2 decimal places for entry into eCampus. Do not enter units. Example: 1.23

Explanation:

First ball has mass M1 = 1 kg

And it's velocity V1i = 10m/s

Mass of second ball M2 = 2kg

Velocity of mass two V2i = 0m/s

Ball one and ball two collides and it is inelastic collision, showing that after collision the two ball stick together and moves with a common velocity.

Ball two now collides with ball three and it is also inelastic collision

Mass of ball three M3 = 0.9kg

Velocity of ball three V3i = 5m/s

Final velocity of ball 1 V1f = 2.5 m/s left

Check attachment for the diagram

Using conservation of momentum

Momentum before collision = Momentum after collision

M1•V1i + M2•0 - M3•V3i = -M1•Vfi + (M2+M3)•V23f

1 × 10 + 0 - 0.9 × 5 = -1 × 2.5 + (2 + 0.9)•V23f

10 - 4.5 = -2.5 + 1.1•V23f

5.5 = -2.5 + 1.1•V23f

1.1•V23f = 5.5+2.5

1.1•V23f = 8

V23f = 8 / 1.1

V23f = 7.27 m/s

To 2d.p

V23f = 7.3m/s

The final velocity of the both mass two and mass three is 7.3 m/s

To input the answer you will write 7.3

neither accurate nor precise.