# Momentum, impulse and projectile motion – problems and solutions

1. A 0.2-kg ball will be inserted into hole C, as shown in the figure below. Hitter strikes the ball in 0.01 second and the path of B-C traveled in 1 second. Determine the magnitude of the force so the ball can be inserted into hole C. Acceleration due to gravity is 10 m/s^{2}.

__Known :__

Angle (θ) = 60^{o}

Mass of ball (m) = 0.2 kg

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

Time interval (Δt) = 0.01 second

Time interval to travel path B-C (t) = 1 second

__Wanted :__ Force (F)

__Solution :__

Equation of impulse : I = F Δt

Equation of the change in momentum : Δp = m (v_{t }– v_{o}).

The impulse equals the change in momentum :

I = Δp

F Δt = m (v_{t }– v_{o})

F = m (v_{t} – v_{o}) / Δt

__Known :__

Δt = 0.01 second

m = 0.2 kg

*v*_{t }*= **the final speed in the equation of impulse-momentum **= **the initial speed of ball **(v*_{o}*) **in the projectile motion*

v_{o} = the initial speed in the equation of impulse-momentum = 0 m/s (initially ball at rest)

F = m (v_{t }– v_{o}) / Δt

F = 0.2 (v_{t }– 0) / 0.01

F = 0.2 v_{t } / 0,01

*continued……*

**Determine the initial speed of the ball ****(v**_{o}**) ****in projectile motion**

Since the ball is hit until the ball reaches the point B = part 1 of the projectile motion.

The ball travels from point B to C = part 2 of the projectile motion.

**Part 2 of the projectile motion :**

The projectile motion could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at a constant velocity and the y motion occurs at a constant acceleration of gravity.

__Known :__

Horizontal distance (x) = 5 meters

Time in air (t) = 1 second

x and t are known so that v_{ox} can calculated using the equation of the uniform linear motion. v_{ox} is the horizontal component of initial speed of ball.

v_{ox} = x / t = 5 meters / 1 second = 5 m/s.

**Part 1 of the projectile motion :**

The horizontal component of speed, v_{ox} is always same so v_{ox }in part 1 of the projectile motion = v_{ox} in part 2 of the projectile motion = 5 m/s.

__Known __

v_{ox} = 5 m/s

θ = 60^{o}

v_{ox }and θ are known so the initial speed (v_{o}) can be calculated.

cos θ = adj / hyp

cos θ = v_{ox} / v_{o}

v_{o }= v_{ox} / cos θ = 5 / cos 60^{o }= 5 / 0.5 = 10 m/s

The initial speed (v_{o}) is 10 m/s.

*The initial speed of the ball **(v*_{o}*) **in projectile motion **= **the final speed of the ball **(v*_{t}*) **in the equation of impulse-momentum.*

**Determine the magnitude of force ****(F)**

F = 0.2 v_{t } / 0.01

F = 0.2 (10) / 0.01

F = 2 / 0.01

F = 200 Newton