A ball having a mass of 0.5 kg is moving towards the east with a speed of 8.0 ms⁻¹. After being hit by a bat it changes its direction and starts moving towards the north with a speed of 6.0 ms⁻¹. If the time of impact is 0.1 s, calculate the impulse and average force acting on the ball.

13 hours ago

Introduction

This problem involves a change in the direction of motion of a ball due to a hit by a bat. Since the direction changes from east to north, we must deal with vector quantities. The concepts of impulse and average force are used here, which are based on the impulse-momentum theorem.

Given Data

Mass of the ball, m = 0.5 kg
Initial velocity, u = 8.0 m/s (east direction)
Final velocity, v = 6.0 m/s (north direction)
Time of impact, t = 0.1 s

Step 1: Represent Initial and Final Velocities as Vectors

Let us consider east as the x-direction and north as the y-direction.

Initial velocity vector, u⃗ = (8, 0) m/s
Final velocity vector, v⃗ = (0, 6) m/s

Step 2: Change in Momentum (Δp⃗)

Change in momentum = Final momentum – Initial momentum

Δp⃗ = m × (v⃗ – u⃗) = 0.5 × [(0 – 8)î + (6 – 0)ĵ]

Δp⃗ = 0.5 × (-8î + 6ĵ) = (-4î + 3ĵ) kg·m/s

Impulse

Impulse = Change in momentum = -4î + 3ĵ kg·m/s

Magnitude of impulse:

|Impulse| = √((-4)² + 3²) = √(16 + 9) = √25 = 5 N·s

Step 3: Average Force

Average Force = Impulse / time

F⃗ = Δp⃗ / t = (-4î + 3ĵ) / 0.1 = -40î + 30ĵ N

Magnitude of average force:

|F| = √(40² + 30²) = √(1600 + 900) = √2500 = 50 N

Step 4: Direction of Average Force

Let θ be the angle the force makes with the east (x-axis):

tan θ = 30 / 40 = 0.75 → θ ≈ arctan(0.75) ≈ 36.87° north of west (since x-component is negative)

Final Answers

Impulse: 5 N·s
Average Force: 50 N
Direction: 36.87° north of west

Conclusion

By applying vector analysis and the impulse-momentum theorem, we find that the bat delivers an impulse of 5 N·s, resulting in an average force of 50 N acting at an angle of approximately 36.87° north of west on the ball.