A steel ball with a mass of 32.0 g is dropped from a height of 2.10 m onto a hor

Question

A steel ball with a mass of 32.0 g is dropped from a height of 2.10 m onto a horizontal steel slab. The ball rebounds to a height of 1.40 m .

Part A

Calculate the impulse delivered to the ball during the impact.

SubmitMy AnswersGive Up

Part B

If the ball is in contact with the slab for 1.90 ms , find the average force on the ball during the impact.

SubmitMy AnswersGive Up

A steel ball with a mass of 32.0 g is dropped from a height of 2.10 m onto a horizontal steel slab. The ball rebounds to a height of 1.40 m .

Part A

Calculate the impulse delivered to the ball during the impact.

p =   kgm/s  

SubmitMy AnswersGive Up

Part B

If the ball is in contact with the slab for 1.90 ms , find the average force on the ball during the impact.

F =   N  

SubmitMy AnswersGive Up

Explanation / Answer

Impulse is the change of momentum. Change of momentum = mass * change of velocity. During the time of the collision, the velocity changes from the final speed downward to the initial speed upward.

For the down trip, the velocity increases 9.8 m/s each second.
Final velocity v 2=u2+2gh
u = 0 m/s

v = sqrt(2 * 9.8 * 2.1) = 6.42 m/s down

For the up trip, the velocity decreases 9.8 m/s each second.
Final velocity up   v = 0 m/s
v 2=u2-2gh

u =sqrt( 2 * 9.8 * 1.4)
u = 5.24 m/s up

Change of velocity v = 5.24 – (-6.42) = 11.66 m/s
Change of momentum = 0.032 * 11.66 = 0.373
Impulse = 0.373 Kg m/s

B) If the ball is in contact with the slab for a time of 1.9 ms-
Force * time = mass * change of velocity

F x t = m x v

F* 1.9 * 10-3 = 0.373
F = 196.32 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.