The instant before a squishy ball of mass meets a horizontal floor and comes to rest in time ?t it has momentum p = p0 (ˆi −ˆj ) .What average force does the floor exert on the ball in order to stop it during this time interval?
it is in magnitude twice momentum, however, not knowing the rebound angle, I don't know which direction (i,r) reverses.
po/t (j-i)
To find the average force exerted by the floor on the ball, we can use the impulse-momentum principle. The impulse-momentum principle states that the change in momentum of an object is equal to the impulse applied to it.
Here's the step-by-step process to calculate the average force exerted by the floor on the ball:
1. Start by finding the initial momentum of the ball (p0) using the given equation p = p0 (ˆi − ˆj). Since the momentum is a vector quantity, it has both magnitude and direction.
- In this case, the mass of the ball is not given, so we can assume it as "m."
Therefore, the initial momentum (p0) can be expressed as p0 = m(ˆi − ˆj).
2. Determine the final momentum of the ball (pfinal). Since the ball comes to rest, the final momentum is zero.
pfinal = 0
3. Calculate the change in momentum of the ball (∆p) using the formula:
∆p = pfinal - p0
In this case, ∆p = 0 - p0 = -p0
4. Calculate the time interval (∆t) which is given as ∆t = t in the question.
5. Apply the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it. Mathematically, it is represented as:
∆p = Favg * ∆t
where Favg represents the average force and ∆t is the time interval.
6. Rearrange the equation to solve for the average force:
Favg = ∆p / ∆t
Since ∆p = -p0, the equation becomes:
Favg = -p0 / ∆t
Substituting the values, the average force exerted by the floor on the ball can be expressed as:
Favg = -p0 / ∆t
Note: The negative sign indicates that the average force is in the direction opposite to the initial momentum vector (ˆi − ˆj).