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Difficulty: EasyNewton's Laws of Motion and Linear Momentum

A tennis ball of mass 0.2 kg0.2\text{ kg} moving horizontally at a speed of 15 m s−115\text{ m s}^{-1} strikes a vertical wall and rebounds along its original path at a speed of 10 m s−110\text{ m s}^{-1}. What is the magnitude of the impulse exerted by the wall on the ball?

  1. 5.0 N s5.0\text{ N s}Answer
  2. B
    1.0 N s1.0\text{ N s}
  3. C
    3.0 N s3.0\text{ N s}
  4. D
    2.0 N s2.0\text{ N s}

Answer

5.0 N s5.0\text{ N s}
The correct answer accounts for the vector nature of velocity. Taking the initial direction as positive (+15 m s−1+15\text{ m s}^{-1}), the rebound velocity is negative (−10 m s−1-10\text{ m s}^{-1}). The change in velocity is Δv=−10−15=−25 m s−1\Delta v = -10 - 15 = -25\text{ m s}^{-1}. Multiplying by mass (0.2 kg0.2\text{ kg}) gives an impulse magnitude of 5.0 N s5.0\text{ N s}.

Step-by-Step Solution

1
Assign direction signs to the initial and final velocity vectors
Initial velocity u=+15 m s−1u = +15\text{ m s}^{-1}, final velocity after rebound v=−10 m s−1v = -10\text{ m s}^{-1}
Velocity is a vector quantity, so reversing direction requires a change in sign.
2
Apply the impulse-momentum theorem (I=Δp=m(v−u)I = \Delta p = m(v - u))
I=0.2×(−10−15)=0.2×(−25)=−5.0 N sI = 0.2 \times (-10 - 15) = 0.2 \times (-25) = -5.0\text{ N s}
Impulse is equal to the net change in linear momentum.
3
Determine the magnitude of the impulse
∣I∣=5.0 N s|I| = 5.0\text{ N s}
Magnitude is the absolute value of the vector quantity.

Key Concept

Impulse-Momentum Theorem in One-Dimensional Rebound Scenarios
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Topics

Newton's Laws of Motion and Linear Momentum

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