Physics, 28.01.2020 20:42 itsme123427

Aball of mass 0.120 kg is dropped from rest from a height of 1.25 m. it rebounds from the floor to reach a height of 0.820 m. what impulse was given to the ball by the floor?

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Answer from: julialombardo53

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)$

$V^{2} = 24.5 m/s$

$V = \sqrt{24.5} \ m/s$

$V = 4.95 \ m/s$

V = ± 4.95 m/s

V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)$

$0 = U^{2} - 16.072 m/s$

$U^{2} = 16.072 m/s$

$U = \sqrt{16.072} \ m/s$

U = ± 4.01 m/s

U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

$F = \frac{mv - mu}{t}$

F.t = m(v - u)

⇒ Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,

∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )

F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )

F.t = 0.120 kg × 8.96 m/s

Impulse = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

Answer from: Quest

50 inches should be the resultant vector

Answer from: Quest

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

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Aball of mass 0.120 kg is dropped from rest from a height of 1.25 m. it rebounds from the floor to r...

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