Impulse and Coefficient of Restitution from Ball Drop

[KatyM7]

A 1.2kg ball dropped from a height of 3m onto a steel plate rigidly attached to the ground bounces back up to a height of 2.5m.(a). What is the impulse delivered to the ball by the plate? (b). What is the coefficient of restitution of the collision?I know my instructor wants me to use something along the lines of a=Δv/Δt; ma= m (Δv/Δt)= Δp/Δt; delta p= maΔt= (ΣF)Δt
Honestly I'm lost on this one because of the whole Δt when I am not given time...

 

[collinsmark]

Hello KatyM7,

Welcome to Physics Forums!

There are two equivalent ways to interpret impulse.

The first, is impulse is force times time: The force applied to an object multiplied by the time interval that the force was applied. (For a uniform force, this is simply $\vec J = \vec F \Delta t$. On the other hand if the force varies with time, then a little calculus is involved, but that's more than I want to get into here.)

That definition is fine and dandy, but it doesn't help you much for this particular problem since you don't know what the force was nor how long it was applied. Fortunately, it's not the only way to view impulse.

For reasons which you've already worked out in your original post, impulse is can also be viewed as an object's change in momentum. What is the ball's momentum at the instant after it collides with the steel plate relative to its momentum at the instant before the collision?

(Hint: don't forget that momentum, like force and impulse, is a vector with both magnitude and direction.)

 

[KatyM7]

Hello KatyM7,

Welcome to Physics Forums!

There are two equivalent ways to interpret impulse.

The first, is impulse is force times time: The force applied to an object multiplied by the time interval that the force was applied. (For a uniform force, this is simply $\vec J = \vec F \Delta t$. On the other hand if the force varies with time, then a little calculus is involved, but that's more than I want to get into here.)

That definition is fine and dandy, but it doesn't help you much for this particular problem since you don't know what the force was nor how long it was applied. Fortunately, it's not the only way to view impulse.

For reasons which you've already worked out in your original post, impulse is can also be viewed as an object's change in momentum. What is the ball's momentum at the instant after it collides with the steel plate relative to its momentum at the instant before the collision?

(Hint: don't forget that momentum, like force and impulse, is a vector with both magnitude and direction.)

The ball's acceleration the instant before the collision would be equal to gravity (9.8m/s^2) and after the collision since it is moving in the opposite direction a=-9.8m/s^2. The ball's mass doesn't change, again the unknown is time. If I am not mistaken this in an isolated system? If so then P(final)= P(initial) in the opposite direction, therefore Δp=0. Please correct me if I am mistaken. Thank you for your help! :)

 

[collinsmark]

The ball's acceleration the instant before the collision would be equal to gravity (9.8m/s^2)

Well, okay. But I was asking more about momentum though.

But yes, the ball acceleration the instant before the collision is 9.8 m/s² in the down direction.

and after the collision since it is moving in the opposite direction a=-9.8m/s^2.

Well, no, that's not true. After the collision, even though the ball is moving up (velocity is up) its acceleration is still 9.8 m/s² in the down direction.

But anyway, I'm asking about the ball's change in momentum. Not the ball's change in acceleration.

Momentum, $\vec p$ is

$$\vec p = m \vec v$$

The ball's mass doesn't change, again the unknown is time. If I am not mistaken this in an isolated system?

The ball is definitely not an isolated system. It encounters the external impulse brought on by the steel plate, and not to mention that it is undergoing an external gravitational force the whole while.

If so then P(final)= P(initial) in the opposite direction, therefore Δp=0. Please correct me if I am mistaken.

I believe that answer is mistaken.

Use your kinematics equations to find the velocity of the ball immediately before impact knowing the initial height of the ball is 3 m. Alternatively, you could use conservation of energy; either way is fine. Then do the same to find the velocity of the ball immediately after the collision, knowing the final height of the ball is 2.5 m.

Finally, calculate the ball's change in momentum during the collision with the steel plate. [Edit: then move on to part b.]

 

[HalfLostAndSearching]

(a). To calculate the impulse delivered to the ball by the plate, we can use the formula Δp = FΔt, where Δp is the change in momentum, F is the average force applied, and Δt is the time interval over which the force acts. In this case, we know that the ball has a mass of 1.2kg and is dropped from a height of 3m onto the steel plate, causing it to bounce back up to a height of 2.5m. We can assume that the collision between the ball and the plate is instantaneous, so the time interval (Δt) is very small. Therefore, we can use the formula Δp = mΔv, where m is the mass of the ball and Δv is the change in velocity. The initial velocity of the ball is 0m/s and the final velocity is -2.5m/s (since the ball bounces back up). Therefore, the change in velocity (Δv) is -2.5m/s - 0m/s = -2.5m/s. Plugging in the values, we get Δp = (1.2kg)(-2.5m/s) = -3kgm/s. This is the impulse delivered to the ball by the plate.

(b). The coefficient of restitution (e) is a measure of the elasticity of a collision. It is defined as the ratio of the final velocity to the initial velocity after a collision. In this case, the initial velocity is 0m/s and the final velocity is -2.5m/s. Therefore, we can calculate the coefficient of restitution as e = (-2.5m/s)/0m/s = 0.833. This means that 83.3% of the ball's initial kinetic energy is retained after the collision with the steel plate.