A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. What impulse was given to the ball by the floor?

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

Since, the motion is downward, final velocity must be negative. So,

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity () = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

Since, the motion is upward, final velocity must be positive. So,

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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