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Difficulty: MediumNewton's Laws of Motion and Linear Momentum

A ball of mass 0.20 kg0.20\text{ kg} moving horizontally towards a vertical wall at a speed of 15 m s−115\text{ m s}^{-1} rebounds in the opposite direction at 10 m s−110\text{ m s}^{-1}. If the impact with the wall lasts for 0.020 s0.020\text{ s}, what is the magnitude of the average force exerted on the ball by the wall?

  1. A
    50 N50\text{ N}
  2. 250 N250\text{ N}Answer
  3. C
    1250 N1250\text{ N}
  4. D
    0.10 N0.10\text{ N}

Answer

The magnitude of the average force exerted on the ball by the wall is 250 N250\text{ N}.
The average force is determined by Newton's second law (F=ΔpΔtF = \frac{\Delta p}{\Delta t}). Because the ball rebounds in the opposite direction, velocity changes from +15 m s−1+15\text{ m s}^{-1} to −10 m s−1-10\text{ m s}^{-1}, yielding a total velocity change magnitude of 25 m s−125\text{ m s}^{-1}. Multiplying by mass (0.20 kg0.20\text{ kg}) gives an impulse magnitude of 5.0 N s5.0\text{ N s}. Dividing impulse by the contact time (0.020 s0.020\text{ s}) yields 250 N250\text{ N}.

Step-by-Step Solution

1
Establish vector direction and assign initial and final velocities
Initial velocity u=+15 m s−1u = +15\text{ m s}^{-1}, final velocity v=−10 m s−1v = -10\text{ m s}^{-1}
Velocity is a vector quantity, so reversing direction requires a opposite sign convention.
2
Calculate the change in momentum (impulse)
\Delta p = m(v - u) = 0.20 \times (-10 - 15) = 0.20 \times (-25) = -5.0\text{ N s}
Impulse is equal to the change in linear momentum.
3
Calculate the magnitude of the average force
F = \frac{|\Delta p|}{\Delta t} = \frac{5.0\text{ N s}}{0.020\text{ s}} = 250\text{ N}
By Newton's second law, average force equals rate of change of momentum (F = \Delta p / \Delta t).

Key Concept

Impulse-Momentum Theorem and Vector Nature of Momentum
Estimated Time:1m 0s
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Topics

Newton's Laws of Motion and Linear Momentum

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