How Does Hitting a Cue Ball Above the Equator Affect Its Motion?

[bluejay1]

Homework Statement

A cue ball is hit 4.5 cm above the horizontal surface uk = 0.35 it is on.
The ball is 6 cm in diameter and has a mass of 250 g. It is hit
by a cue stick with an average force of
6000 N during a contact time of 20 usec.

(a) What is the ball’s speed after it is hit?

Homework Equations

F=Ma
x=x0+v0xt+1/2at^2

The Attempt at a Solution

I can't figure out the effect being hit above the equator has and how to incorporate it?
Thanks

 

[turin]

It sounds like they want you to take into account the torque on the cue ball due to the fact that it was hit off center. I guess you assume that the cue does not slip and that it hits parallel to the horizontal surface, but it doesn't actually say that, so I'm just guessing.

 

[thomate1]

for your question. I would approach this problem by first breaking it down into smaller, more manageable parts. Let's start with the basic equations of motion, which you have already listed:

F=Ma (Newton's Second Law)
x=x0+v0xt+1/2at^2 (Equation of motion)

We also need to consider the coefficient of kinetic friction, uk, which is given as 0.35 in this problem. This value represents the amount of friction between two surfaces in motion. In this case, it is the friction between the cue ball and the surface it is rolling on.

Next, we need to think about the initial conditions of the cue ball. It is hit 4.5 cm above the horizontal surface, which will affect its initial velocity and trajectory. We also know the diameter and mass of the ball, which will be important for calculating its final speed.

Now, let's go through the steps to solve this problem:

1. Determine the initial velocity: Since the cue ball is hit by a cue stick with an average force of 6000 N during a contact time of 20 usec, we can use the equation F=Ma to calculate the initial acceleration of the ball. We know the mass of the ball is 250 g, so we can rearrange the equation to solve for acceleration: a=F/m = 6000 N / 0.25 kg = 24000 m/s^2. Now, we can use the equation of motion, x=x0+v0xt+1/2at^2, to solve for the initial velocity (v0) of the ball.

2. Account for the friction: Since the ball is rolling on a surface, we need to take into account the friction between the ball and the surface. We can do this by using the equation Ff=ukFn, where Ff is the force of friction, uk is the coefficient of kinetic friction, and Fn is the normal force (equal to the weight of the ball, mg). We can then subtract this force from the initial force (F) to get the net force acting on the ball.

3. Calculate the final speed: Now that we have the initial velocity and the net force acting on the ball, we can use the equation x=x0+v0xt+1/2at^2 to solve for the final velocity (v) of the ball.

By following these steps, we can calculate