**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

**What is the relationship between impulse and change in momentum?****Answer:**Impulse is the product of force and the time over which it acts, and it is equal to the change in momentum of an object. Mathematically, $Impulse=Δp$, where $Δp$ is the change in momentum.

**How does an increase in the time of impact, such as with a crumple zone in a car, affect the force experienced during a collision?****Answer:**Increasing the time of impact reduces the average force experienced during a collision. This is because impulse (change in momentum) is constant for a given collision, and by increasing time, the force is spread out over a longer duration, hence decreasing the average force.

**What remains constant for a projectile in motion under the influence of gravity alone?****Answer:**The horizontal velocity of a projectile remains constant when only gravity acts on it. The vertical velocity, however, changes due to gravitational acceleration.

**Why is the path of a projectile under the influence of gravity parabolic?****Answer:**The path is parabolic because while the horizontal velocity remains constant, the vertical velocity is constantly changed by gravitational acceleration, leading to a quadratic relationship between horizontal and vertical displacements.

**What happens to the momentum of an isolated system if no external forces act on it?****Answer:**If no external forces act on an isolated system, its total momentum remains conserved. This is known as the conservation of momentum.

**How does the impulse provided by a baseball bat to a ball compare when the ball is hit for a home run versus when it’s just tapped lightly?****Answer:**The impulse provided to the baseball is greater when hit for a home run compared to when it’s tapped lightly, as the change in momentum of the ball (from stationary to flying off the bat) is much greater in the home run scenario.

**If a projectile is launched from ground level at an angle and returns to ground level, how do its launch and landing speeds compare?****Answer:**Assuming no air resistance, the projectile’s landing speed will be equal to its launch speed. This is due to the conservation of energy.

**How do the launch angle and range of a projectile relate, assuming no air resistance?****Answer:**For a given initial speed, the maximum range of a projectile is achieved at a launch angle of 45 degrees. Launching at angles less or greater than 45 degrees will result in a shorter range.

**Why does increasing the time over which a force acts on an object, such as catching an egg with a slight hand movement, reduce the likelihood of breaking or damaging the object?**

**Answer:**Increasing the time over which a force acts spreads the impulse (change in momentum) over a longer duration, thereby reducing the average force experienced by the object. By catching an egg with a slight hand movement, the force of the catch is spread out, reducing the chance of the egg breaking.