After falling from rest from a height of 27 m, a 0.53 kg ball rebounds upward, reaching a height of 17 m. If the contact between ball and ground lasted 1.6 ms, what average force was exerted on the ball?
Vf = sqrt (2 * 9.8* 27m) = 23 m/s
Vi = sqrt (2 * 9.8* 17m) = 18.25 m/s
a = change in velocity/ change in time
= Vf - Vi/ tf - ti
= [23 - (-18.25)] / [.0016 s - 0s]
= 41.25 / .0016 = 25781.25 m/s^2
you know F = ma
so F = .53kg * 25781.25 m/s^2
F = 13664 N
The reason why the velocity is negative in after the bounce I think is because it is now moving downward.
The answer is correct though for sure.
Had the same problem to work on :p
ACTUALLY IT IS BECAUSE THE FINAL VELOCITY BECOMES THE INITIAL VELOCITY IN THE SECOND ECUATION AND VICEVERSA, SO IT SHOULD BE [18.25 - (-23)] / [.0016 s - 0s] , BECAUSE THE NEGATIVE V IS 23...
This is wrong!
This was correct for me, thank you!
How did you guys find time?
It gives you time, convert 1.6 ms to seconds.
To find the average force exerted 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 force applied to it multiplied by the time it takes for the force to be applied.
First, let's find the initial velocity of the ball just before it hits the ground. We can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
In this case, the ball falls from rest, so the initial velocity (u) is 0 m/s. The distance traveled (s) is 27 m, and the final velocity (v) just before hitting the ground can be found using the same equation by substituting g (acceleration due to gravity) as the acceleration and taking the negative sign because the velocity is in the opposite direction of the distance traveled.
v^2 = u^2 + 2as
v^2 = 0 + 2(-g)(-27)
v^2 = 2g(27)
v = √(2g(27))
Next, let's find the final velocity of the ball just after it rebounds. The final velocity (u') just after rebounding is the negative of the final velocity just before hitting the ground (v) because the ball changes direction.
u' = -v
Now we can calculate the change in momentum (Δp), which is the change in velocity (Δv) multiplied by the mass (m) of the ball:
Δp = m(u' - u)
Finally, we can find the average force (F) by dividing the change in momentum (Δp) by the time (Δt) it takes for the force to be applied:
F = Δp / Δt
In this case, the time (Δt) is given as 1.6 ms, which is 1.6 × 10^-3 s.
By plugging in the values and following these steps, you can calculate the average force exerted on the ball.