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lorasvet [3.4K]

2 years ago

10

A ball of mass m is dropped from rest from a height h and collides elastically with the floor, rebounding to its original height

.
What is the magnitude of the average force applied by the floor on the ball during the time the ball is in contact with the floor?

(A) Zero
(B) mg
(C) 2mg
(D) 4mg
(E) It cannot be determined without knowing the length of time that the ball is in contact with the floor.

2 answers:

grin007 [14]2 years ago

5 0

Answer:

option (E)

Explanation:

mass = m

height = h

initial velocity u = 0

The average force applied by the floor on the ball is the rate of change of momentum of the ball.

So, we need the time of contact of ball and floor, so it is not given in the question.

Thus, option (E)

Jet001 [13]2 years ago

3 0

Answer:

Explanation:

Given

mass m collides elastically with floor and jump to its original height I .e.velocity of rebound is same as the initial velocity

Impulse imparted by steel ball to the ground is given by the change in momentum of steel ball

$J=\Delta P$

and impulse is product of average force and time

so we need time to calculate the average force. So, option e is correct

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Answer:

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No.

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3 years ago

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iVinArrow [24]

Answer:

Explanation:

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Gravitational potential energy = mass*acc due to gravity * height

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v = velocity of the body (in m/s²)

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Since mass = 3.8kg

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