Rick is playing billiards and hits the cue ball. It makes contact with the 8 ball head on. If before the collision the cue ball was travelling at 2.4 m/s and after the collision the 8 ball was traveling at 2.13 m/s. How fast and in what direction is the cue ball traveling? (Mass of cue ball=.17kg and mass of 8 ball=.16kg)
---I found the velocity of the cue ball to be 7.04 m/s but I don't know how to find the direction.. Help and thx!!!
since it hit head-on, the balls travel in exactly opposite directions, or in the same direction.
Aside from that, since the balls are of almost identical mass, do you really expect one of them to take off at 3 times the original speed?
to conserve momentum,
.17(2.4) = .17v + .16(2.13)
v = .395
That is, the cue ball almost came to a stop, and virtually all of its momentum was transferred to the 8-ball. The cue ball is still traveling slowly in its original direction.
Head on means the ball stays on the same path, or reverse. Which? use algebra, and the law of conservation of moentum
momentum before=momentum after
M*2.4=MV1+MV2
= MV1+2.13M
V1= 2.4-2.13 which is positve, in the same direction it started in.
Okay, so i did my math wrong sorry and thx Steve!! and bobpursley, does that mean the cue ball is traveling .27 m/s straight?? because usually they want angle but since it's head on it's just straight rite?
To find the direction in which the cue ball is traveling after the collision, we need to calculate the velocity vector. Since we are given only the magnitude of the velocity (2.13 m/s), we need another piece of information to determine the direction.
One way to determine the direction is by considering the conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object can be calculated by multiplying its mass by its velocity. Let's denote the velocity of the cue ball after the collision as v_cue and the velocity of the 8 ball after the collision as v_8.
Using the conservation of momentum, we have:
(mass of cue ball) * (velocity of cue ball before collision) + (mass of 8 ball) * (velocity of 8 ball before collision) = (mass of cue ball) * (velocity of cue ball after collision) + (mass of 8 ball) * (velocity of 8 ball after collision)
(0.17 kg) * (2.4 m/s) + (0.16 kg) * (0 m/s) = (0.17 kg) * (v_cue) + (0.16 kg) * (2.13 m/s)
After simplifying the equation, we can solve for v_cue:
(0.17 kg) * (2.4 m/s) = (0.17 kg) * (v_cue) + (0.16 kg) * (2.13 m/s)
0.408 kg·m/s = 0.17 kg·v_cue + 0.3392 kg·m/s
0.408 kg·m/s - 0.3392 kg·m/s = 0.17 kg·v_cue
0.0688 kg·m/s = 0.17 kg·v_cue
v_cue = (0.0688 kg·m/s) / (0.17 kg)
v_cue ≈ 0.405 m/s
So, the velocity of the cue ball after the collision is approximately 0.405 m/s. To determine the direction, we need additional information or assumptions about the motion of the balls (e.g., the angle of impact, any spin on the balls, etc.). Without this information, we cannot determine the exact direction of the cue ball.