A 2.00-kilogram mud ball drops from rest at a height of 17.0 m. If the impact between the ball and the ground lasts 0.46 s, what

Question

3 years ago

5

is the magnitude of the average force exerted by the ball on the ground

1 answer:

Softa [21] · Answer 1 · 2022-06-29T11:05:50+03:00

Answer: 79.35 N

Explanation: according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formulae below

v² = u² + 2gh

Where h = height of the cliff = 17m

v² = 2 × 9.8 × 17

v² = 333.2

v = √333.2

v = 18.25 m/s

At t = 0.46s and v = 18.25 m/s, we can get the average force of impact

F×0.46 = 2 (18.25 - 0)

F × 0.46 = 2 (18.25)

F × 0.46 = 36.5

F = 36.5 /0.46 = 79.35 N