# Momentum and impulse – problems and solutions

1. A small ball is thrown horizontally with a constant speed of 10 m/s. The ball hits the wall and reflected with the same speed. What is the change in linear momentum of the ball?

__Known :__

Mass (m) = 0.2 kg

Initial speed (v_{o}

Final speed (v_{t}) = 10 m/s

*The plus and minus sign indicates that the objects moves in opposite direction.*

__Wanted ____:__ the change in linear momentum (Δp)

__Solution :__

**Formula of the change in linear momentum **:

Δp = m v_{t} – m v_{o }= m (v_{t} – v_{o})

The change in linear momentum :

Δp = 0.2 (10 – (-10)) = 0.2 (10 + 10)

Δp = 0.2 (20)

**Δp ****= 4 kg m/s**

2. A 10-gram ball falls freely from a height, hits the floor at 15 m/s, then reflected upward at 10 m/s. Determine the impulse!

__Known :__

Mass (m) = 10 gram = 0.01 kg

Initial velocity (v_{o}) = -15 m/s

Final velocity (v_{t}) = 10 m/s

Wanted : Impulse

__Solution :__

The impulse (I) equals the change in momentum (Δp)

I = m v_{t }– m v_{o} = m (v_{t} – v_{o})

Impulse :

I = 0.01 (10 – (-15)) = 0.01 (10 + 15)

I = 0.01 (25)

**I = 0.25 kg m/s**

3. A 200-gram ball thrown horizontally with a speed of 4 m/s, then the ball was hit in the same direction. The duration of the ball in contact with the bat is 2 milliseconds and the ball speed after leaving the bat is 12 m/s. The magnitude of force exerted by the batter on the ball is …

__Known :__

Mass (m) = 200 gram = 0.2 kg

Initial velocity (v_{o}) = 4 m/s

Final velocity (v_{t}) = 12 m/s

Time interval (t) = 2 milliseconds = (2/1000) seconds = 0.002 seconds

__Wanted__ : The magnitude of the force (F)

__Solution :__

Formula of impulse :

I = F t

Formula of the change in momentum :

m v_{t} – m v_{o} = m (v_{t} – v_{o})

The impulse (I) equals the change in momentum (Δp)

I = Δp

F t = m (v_{t} – v_{o})

F (0.002) = (0.2)(12 – 4)

F (0.002) = (0.2)(8)

F (0.002) = 1.6

F = 1.6 / 0.002

F = 800 Newton

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