Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial velocity of the ball A is 6 meters per second to the right, and the initial velocity of the ball B is 12 meters per second to the left.

Question

1. When the two balls hit each other, what will happen if it is a perfectly inelastic collision?
2. Compare the final velocity of the balls.
3. What can you say about the total momentum before and after the collision?

Answer 1

initial momentum to right = 7 * 6 - 2 * 12 = 18 kg m/s
So 18 will be the final momentum to the right as well
so if u is ball a and v is ball b
18 = 7 u + 2 v
now about kinetic energy
initial =(1/2) 7 (36) + (1/2)(2)(144) = 126+72 = 198 Joules
since we said inelastic final = initial so
(1/2) 7 u^2 + (1/2)(2) v^2 = 198
well v= (18-7u)/2
3.5 u^2 + (1/4)(324 - 252 u + 49 u^2) = 198
14 u^2 + 49 u^2 -252u -468 = 0
63 u^2 - 252 u - 468 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
u = 5.38 or -1.38
which?
if -1.38 then v = (18-7u)/2 = (18 + 9.66)/2 = 13.83 possible
if 5.38 then v = (18 -37.66)/2 =- 9.33 possible only if second ball passed right through the first one
so u = -1.38 and v = 13.83

Newton's very first said the momentum did not change. That is what we used.

Answer 2

Given: M1 = 7kg, V1 = 6m/s.
M2 = 2kg, V2 = -12m/s.
V3 = velocity of M1 and M2 after collision.

Momentum before = Momentum after.
M!*V1 + M2*V2 = M1*V3 + M2*V3.
7*6 + 2*(-12) = 7*V3 + 2*V3
9V3 = 18
V3 = 2 m/s.

Answer 3

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