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Difficulty: HardMomentum and Collisions

A cargo crate has a weight of 4,900 N4,900 \text{ N} on Earth (where g=9.8 m/s2g = 9.8 \text{ m/s}^2). It is sliding across a frictionless floor at a velocity of 12 m/s12 \text{ m/s} when it hits a padded wall. If the crate comes to a complete stop in 0.6 s0.6 \text{ s}, what is the average impact force exerted on the crate during this collision?

  1. A
    98,000 N98,000 \text{ N}
  2. 10,000 N10,000 \text{ N}Answer
  3. C
    3,600 N3,600 \text{ N}
  4. D
    25 N25 \text{ N}

Answer

The average impact force is 10,000 N10,000 \text{ N}.
To find the average impact force, we use the impulse-momentum theorem (FΔt=mΔvF\Delta t = m\Delta v) or Newton's second law (F=maF = ma). First, we must convert the weight to mass by dividing by the acceleration due to gravity: 4,900 N/9.8 m/s2=500 kg4,900 \text{ N} / 9.8 \text{ m/s}^2 = 500 \text{ kg}. Next, we calculate the acceleration by dividing the change in velocity by the impact time: 12 m/s/0.6 s=20 m/s212 \text{ m/s} / 0.6 \text{ s} = 20 \text{ m/s}^2. Multiplying the mass by the acceleration gives the force: 500 kg×20 m/s2=10,000 N500 \text{ kg} \times 20 \text{ m/s}^2 = 10,000 \text{ N}.

Step-by-Step Solution

1
Calculate the mass of the crate from its weight.
m=Wg=4,900 N9.8 m/s2=500 kgm = \frac{W}{g} = \frac{4,900 \text{ N}}{9.8 \text{ m/s}^2} = 500 \text{ kg}
Newton's laws and momentum calculations require mass, not weight.
2
Calculate the crate's change in velocity.
Δv=12 m/s−0 m/s=12 m/s\Delta v = 12 \text{ m/s} - 0 \text{ m/s} = 12 \text{ m/s}
The crate comes to a complete stop, so the total change in velocity is its initial velocity.
3
Calculate the crate's acceleration (or deceleration) during the impact.
a=ΔvΔt=12 m/s0.6 s=20 m/s2a = \frac{\Delta v}{\Delta t} = \frac{12 \text{ m/s}}{0.6 \text{ s}} = 20 \text{ m/s}^2
Acceleration is the change in velocity divided by the time it takes to change.
4
Calculate the average impact force using Newton's Second Law.
F=m×a=500 kg×20 m/s2=10,000 NF = m \times a = 500 \text{ kg} \times 20 \text{ m/s}^2 = 10,000 \text{ N}
Force equals mass times acceleration. (Alternatively, impulse FΔt=ΔpF\Delta t = \Delta p yields the same result: F=500×120.6=10,000 NF = \frac{500 \times 12}{0.6} = 10,000 \text{ N}).

Key Concept

The impulse-momentum theorem and the distinction between mass and weight.
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Topics

Momentum and Collisions

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