1. A 0.5 kg ball free fall from a height of h_{1} = 7.2 meters and reflected a height of h_{2} = 3.2 meters. Acceleration due to gravity = 10 m/s^{2}. Determine impulse.

__Known :__

Mass of ball (m) = 0.5 kg

First height (h_{1}) = 7.2 meter

Second height (h_{2}) = 3.2 meter

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ Impulse (I)

__Solution :__

__Velocity of ball before collisions ____(v___{o}__)__

Velocity of ball before collision calculated using equation of free fall motion. __Known :__ Height (h) = 7.2 meters, acceleration due to gravity (g) = 10 m/s^{2}. __Wanted :__ Final velocity after collision.

v^{2} = 2 g h

v_{o}^{2} = 2(10)(7.2) = 144

v_{o} = 2(10)(7.2) = 12 m/s

Velocity of ball before collision (v_{o}) = -12 m/s. Minus sign indicates the direction of ball.

__Velocity of ball after collision ____(v___{t}__)__

Velocity of ball after collision calculated using equation of vertical motion. __Known __: height (h) = 3.2 meters, acceleration due to gravity (g) = -10 m/s^{2}, final velocity at the maximum height (v_{t}^{2}) = 0. __Wanted :__ Initial velocity after collision between ball and floor.

v_{t}^{2} = v_{o}^{2} + 2 g h

0 = v_{o}^{2} + 2 (-10)(3.2)

v_{o}^{2} = 64

v_{o} = √64 = 8 m/s

Velocity of ball after collision (v_{t}) is 8 m/s

__Impulse (I)__

Impulse (I) = the change in momentum (Δp)

I = m (v_{t }– v_{o}) = (0.5)(8-(-12)) = (0.5)(8 + 12) = (0.5)(20) = 10 Newton second

[irp]

2. A 5-gram ball free fall from a height and strikes the floor. Acceleration due to gravity, g = 10 ms^{-2}. Velocity of ball before collision is 6 ms^{-1} and after collision, the ball is reflected upright at 4 m/s. Determine the impulse.

__Known :__

Mass of ball (m) = 5 gram = 0.005 kg

Velocity of ball before collision (v_{o}) = -6 m/s

Velocity of ball after collision (v_{t}) = 4 m/s

Plus and minus sign indicates that the direction before collision is opposite with the direction after collision.

__Wanted:__ Impulse (I)

__Solution :__

Impulse (I) = the total change in momentum (Δp).

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

I = (0.005)(4 – (-6)) = (0.005)(4 + 6) = (0.005)(10) = 0.05 Newton second

[irp]

3. A 20-gram thrown with velocity of 4 m.s^{-1} to the left. After colliding with the wall, the ball is reflected with velocity of v_{2 }= 2 m.s^{-1} to the right. Determine the impulse.

__Known :__

Mass of ball (m) = 20 gram = 0.020 kg

Velocity of ball before collision (v_{o}) = -4 m/s (to the left)

Velocity of ball after collision (v_{t}) = +2 m/s (to the right)

Plus and minus sign indicates the opposite direction.

__Wanted :__ Impulse

__Solution :__

Impulse (I) = the change in momentum (Δp) = m v_{t} – m v_{o}

Impulse (I) = m (v_{t} – v_{o}) = 0.02 (2 – (-4))

Impulse (I) = 0.02 (2 + 4) = 0.02 (6)

Impulse (I) = 0.12 Newton second.