# Momentum – problems and solutions

1. 0.5-kg ball moves at 2 m/s. Then the ball is hit with force F opposite to the ball direction, so the ball speed is changed to 6 m/s. The ball in contact with a hitter for 0.01 second, what is the change in momentum of the ball.

__Known :__

Mass of ball (m) = 0.5 kg

Initial velocity (v_{o}) = 2 m/s

Final velocity (v_{t}) = -6 m/s

Time interval (t) = 0.01 second

__Wanted :__ The change in momentum

__Solution :__

∆p = m v_{t} – m v_{o} = m (v_{t }– v_{o})

∆p = (0.5 kg)(- 6 m/s – 2 m/s)

∆p = (0.5 kg)(-8)

∆p = 4 kg m/s

2. A 100-gram object moves at 5 m/s. Force F works for 0.2 seconds to stop the object. Determine magnitude of force F.

__Known :__

Mass of object (m) = 100 gram = 100/1000 = 0.1 kg

Initial velocity (v_{o}) = 5 m/s

Final velocity (v_{t}) = 0

Time interval (Δt) = 0.2 seconds

__Wanted :__ Magnitude of force F

__Solution :__

I = ΔP

F (Δt) = m (v_{t} – v_{o})

F (0.2) = 0.1 (0 – 5)

F (0.2) = 0.1 (– 5)

F (0.2) = -0,5

F = -0,5 / 0.2

F = -2.5 N

Magnitude of force F = 2.5 Newton. Minus sign indicates the direction of force opposite with the direction of the object.

3. A 100-gram ball free fall from the height of 20 cm without initial velocity and then the ball hits the floor. After collision, the ball is reflected upward (acceleration due to gravity is 10 ms^{-2}

__Known :__

Mass of ball (m) = 100 gram = 0.1 kg

Height (h) = 20 cm = 0.2 meters

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

Velocity of ball after hits the floor (v_{t}) = 1 m/s

__Wanted :__ The change in momentum

__Solution :__

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

Calculate velocity of ball before collision using equation of free fall motion. __Known :__ height of ball (h) = 0.2 meters, acceleration due to gravity (g) = 10 m/s^{2}. __Wanted :__ velocity of ball when hits the floor.

v^{2} = 2 g h

v^{2} = 2 (10)(0.2) = 4

v = √4 = -2 m/s

Minus sign indicates that the direction of ball before collision is opposite with the direction of ball after collision.

__The change in momentum of ball (Δp) __

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

Δp = (0.1)(1 – (-2)) = (0.1)(1 + 2) = (0.1)(3) = 0.3 Newton second

4. An object initially at rest, explosion into 2 parts with a ratio of 3 : 2. The larger part of the mass is thrown at a speed of 20 m/s. What is the velocity of the smaller part.

__Known :__

Mass of object 1 before explosion = m

Velocity of object 1 before explosion = 0 (object at rest)

Mass of the larger part after explosion (m_{1}) = 3m

Mass of the smaller part after explosion (m_{2}) = 2m

Velocity of the larger part after explosion (v_{1}‘) = 20 m/s

__Wanted :__ Velocity of the smaller part after explosion (v_{2}‘)

__Solution :__

The equation of the law of the conservation of momentum :

m_{1 }v_{1 }= m_{1 }v_{1}‘ + m_{2 }v_{2}‘

(m)(0) = (3m)(20) + (2m) v_{2}‘

0 = 60m + (2m) v_{2}‘

60m = -2m v_{2}‘

60 = -2 v_{2}‘

v_{2}‘ = -60/2

v_{2}‘ = -30 m/s

Minus sign indicates that the direction of the smaller part is opposite with the direction of the larger part.