Momentum Impulse and Projectile motion – Problems and Solutions

1 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/s².

Known :

Angle (θ) = 60^o

Mass of ball (m) = 0.2 kg

Acceleration due to gravity (g) = 10 m/s²

Time interval (Δt) = 0.01 second

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

Wanted : Force (F)

Solution :

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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

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