a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landing?

Answers

Answer 1

The change in momentum is Δp= 0.028 kg m/s
Explanation:
An impulse describes a change in momentum. The change in momentum of an object is its mass times the change in its velocity.
The change in moment is given by the expression below
Δp=m⋅(Δv)=m⋅(vf−vi) .
Given data
mass m= 0.140-kg
initial velocity vi= 120 m/s
final velocity vf= 1 m/s
substituting we have
Δp=m⋅(Δv)=0.14⋅(1−1.2)
Δp=m⋅(Δv)=0.14⋅(-0.2)
Δp= 0.028 kg m/s
The change in momentum was found to be Δp= 0.028 kg m/s

Answer 2

The speed of the ball (i.e the velocity since it has a direction in a specified time) is 5.2 m/s before it landed.

From the given information,

the mass of the baseball = 0.15 kgthe momentum of the baseball = 0.78 kg m/s

For an object moving in a direction in a specified time, the momentum of the object is usually directly proportional to the mass and velocity.

∴

Momentum = mass × velocity

velocity = momentum/mass

velocity = 0.78 kgm/s ÷ 0.15 kg

velocity = 5.2 m/s

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Answers

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Answers

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