Math Problem Statement

A tennis ball strikes a vertical wall as shown in the overhead view. The 142 g ball has a speed of 6.5 m/s and strikes the wall at an angle of θ =40∘ and then rebounds with the same speed and angle. It is in contact with the wall for 4.1 ms.

What is the magnitude of the impulse on the ball?

Solution

To find the magnitude of the impulse JJ exerted on the ball by the wall, we’ll use the definition of impulse:

J=ΔpJ = \Delta p

where Δp\Delta p is the change in momentum. Since the ball rebounds with the same speed and angle, the impulse only affects the component of velocity perpendicular to the wall. Let's go through the steps:

1. Find the Initial and Final Momentum Components

Given:

  • Mass of the ball, m=142g=0.142kgm = 142 \, \text{g} = 0.142 \, \text{kg}
  • Initial speed, v=6.5m/sv = 6.5 \, \text{m/s}
  • Angle of incidence and rebound, θ=40\theta = 40^\circ
  • Contact time, Δt=4.1ms=0.0041s\Delta t = 4.1 \, \text{ms} = 0.0041 \, \text{s}

The initial momentum perpendicular to the wall is: pinitial, perpendicular=mvsinθ=0.142×6.5×sin40p_{\text{initial, perpendicular}} = m \cdot v \cdot \sin \theta = 0.142 \times 6.5 \times \sin 40^\circ

Calculating this:

pinitial, perpendicular=0.142×6.5×0.64280.5936kgm/sp_{\text{initial, perpendicular}} = 0.142 \times 6.5 \times 0.6428 \approx 0.5936 \, \text{kg} \cdot \text{m/s}

Since the ball rebounds with the same speed and angle, the final momentum perpendicular to the wall is in the opposite direction, so:

pfinal, perpendicular=0.5936kgm/sp_{\text{final, perpendicular}} = -0.5936 \, \text{kg} \cdot \text{m/s}

2. Calculate the Change in Momentum (Impulse)

The impulse is the change in momentum in the perpendicular direction:

Δp=pfinal, perpendicularpinitial, perpendicular=0.59360.5936=1.1872kgm/s\Delta p = p_{\text{final, perpendicular}} - p_{\text{initial, perpendicular}} = -0.5936 - 0.5936 = -1.1872 \, \text{kg} \cdot \text{m/s}

The magnitude of the impulse is therefore:

J=1.1872kgm/s|J| = 1.1872 \, \text{kg} \cdot \text{m/s}

Answer

The magnitude of the impulse on the ball is approximately:

J=1.19kgm/s|J| = 1.19 \, \text{kg} \cdot \text{m/s}

Would you like more detailed steps, or do you have any questions about the solution?


  1. How would the impulse change if the ball’s mass increased?
  2. What if the angle of incidence was closer to the wall?
  3. How would a change in speed affect the impulse?
  4. What is the role of contact time in calculating force?
  5. How does impulse relate to changes in momentum in real-world sports?

Tip: Always keep units consistent, especially for momentum and impulse, for accurate results.

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Math Problem Analysis

Mathematical Concepts

Impulse and Momentum
Trigonometry

Formulas

Impulse formula: J = Δp
Momentum calculation: p = m * v
Perpendicular momentum component: p_perpendicular = m * v * sin(θ)

Theorems

Conservation of Momentum

Suitable Grade Level

Grade 11-12