Question

1. On a smooth, frictionless table, a billiard ball of velocity v is moving toward two other aligned billiard balls in contact. What will be the velocity of each ball after impact? Assume that all balls have the same mass and that the collisions are elastic. Ignore any rotation of the balls (Hint: Treat this as two successive collisions.)

2. Repeat the Problem #1 but assume that the middle ball has twice the mass of each of the others

Answer 1

1. Divide this event into two different ones: interaction 1 between the ball 1 and 2, and interaction 2 between the balls 2 and 3 since these balls are different bodies not glued to one another. So, according to law of conservation of momentum and law of conservation of energy:

$$v_1=v, u_1$$

- velocities of the ball 1 before and after interaction,

$$v_2=0, u_2$$

- velocities of the ball 2 before and after interaction.

$$mv_1+mv_2=mu_1+mu_2,$$ $$\frac{mv_1^2}{2}+\frac{mv_2^2}{2}=\frac{mu_1^2}{2}+\frac{mu_2^2}{2},$$

rewrite it:

$$v_1-u_1=u_2-v_2,(1)$$ $$v_1^2-u_1^2=u_2^2-v_2^2, (2)$$

divide (2) (using difference of two squares formula) by (1) and you get:

$$v_1+u_1=u_2+v_2,$$

solve it with (1). The answer is:

$$v_1=u_2, v_2=u_1,$$

which means that

$$u_2=v,u_1=0,$$

or the ball 2 moves with

$$v$$

after interaction and the ball 1 stops.

By analogy for the second interaction between the balls 2 and 3 we can write absolutely the same but change the indexes:

1->2, 2->3:

$$v_2=u_3, v_3=u_2,$$ $$u_3=v,u_2=0,$$

or the ball 3 moves with

$$v$$

after collision and the ball 2 stops.

All in all, in our case with very small distance between the balls 2 and 3 we'll watch the following: the first ball hits the other two resting balls, then the balls 1 and 2 will be resting in contact and the ball 3 will move with the same speed as of the first ball.

2. In this case the overall idea is the same, but remember that

$$v_2=0$$

.

$$mv_1=mu_1+2mu_2,$$ $$\frac{mv_1^2}{2}=\frac{mu_1^2}{2}+\frac{2mu_2^2}{2},$$ $$v_1-u_1=2u_2,(1)$$ $$v_1^2-u_1^2=2u_2^2, (2)$$

divide (2) by (1):

$$v_1+u_1=u_2,$$

solve it with (1). The answer is:

$$u_2=\frac{2}{3}v_1=\frac{2}{3}v,$$ $$u_1=-\frac{v}{3}$$

or after the interaction between 1 and 2 the ball 1 goes backwards, the ball 2 moves with 2/3 of the speed of the first ball.

Now the second collision between 2 and 3. To make it easy, assume that velocity of 2 before the collision is

$$y=2/3\cdot v$$

, after - x, velocity of 3 before collision is 0, after - z:

$$2my=2mx+mz$$ $$2my^2/2=2mx^2/2+mz^2/2,$$

which gives

$$2y-2x=z,$$ $$2y+2x=z.$$

We see that the ball 2 stops, the third ball moves forward with

$$4v/3$$

.

So for the 2nd problem after all the balls will move like that:

ball 1: backwards with

$$v/3$$

;

ball 2: zero velocity;

ball 3: forward with

$$4v/3$$

.

Not so obvious as in the previous problem.