Cue ball at rest on a frictionless table

[rabid_baboon]

A cue ball (mass=0.165kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of +1.50Ns (Newton-seconds) to the ball. The ball the slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

I need help at least getting started.

 

[Bill Foster]

$$Ft=mv$$

 

[ogb p]

To solve this problem, we can use the principle of conservation of momentum. This states that in a closed system, the total momentum before an event must be equal to the total momentum after the event.

In this scenario, the two balls and the cue stick can be considered as a closed system. Before the collision, the cue ball is at rest, so its momentum is zero. The second ball is also at rest, so its momentum is also zero. Therefore, the total momentum before the collision is zero.

After the collision, the cue ball will have a non-zero velocity and the second ball will also have a velocity. However, the total momentum of the system must still be zero. This means that the momentum of the cue ball must be equal in magnitude but opposite in direction to the momentum of the second ball.

We can use the equation for momentum, p=mv, to solve for the velocity of the second ball. Since the cue ball has a mass of 0.165kg and a momentum of +1.50Ns, we can rearrange the equation to solve for its velocity. This gives us v = p/m = (+1.50Ns) / (0.165kg) = +9.09 m/s.

Since the second ball has the same mass and an equal but opposite momentum, its velocity will also be +9.09 m/s. This is the final velocity of the second ball just after it is struck by the cue ball.

It is important to note that this calculation assumes that the collision is perfectly elastic, meaning that there is no loss of kinetic energy during the collision. In real-world scenarios, there may be some energy lost due to friction or other factors, which could affect the final velocity of the second ball.