Question

A rubber ball of mass 0.175 kg is released from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.630 m. What impulse was applied to the ball by the floor? (You may safely ignore air resistance. Enter the magnitude in kg · m/s)

Answer #1

Impulse is defined as change in momentum.

I = M*V1- M*V2

where V1 is the velocity of ball before hitting the floor and V2 after hitting the floor.

and Velocity during free fall is given by,

V = √(2*g*h)

Now calculate V1 and V2,

V1 = √(2*9.81*1.25) = 4.95 m/s

V2 = √(2*9.81 * 0.63) = 3.51 m/s

as V2 have opposite direction of V1 then we will take value of V2 as negative.

I = M * (4.95 - (-3.51))

I = 0.175 * 8.46

I = 1.48 Kg.m/s

**Final Answer :-**

**Impulse = 1.48 kg.m/s**

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