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Difficulty: MediumMomentum and Collisions

A test sled has a weight of 2,940 N2,940 \text{ N} (where g=9.8 m/s2g = 9.8 \text{ m/s}^2). During a braking test, a friction track applies a constant drag force of 588 N588 \text{ N}. If it takes exactly 10 seconds10 \text{ seconds} for the sled to come to a complete halt, what was its initial velocity?

  1. A
    1.96 m/s1.96 \text{ m/s}
  2. B
    2 m/s2 \text{ m/s}
  3. 19.6 m/s19.6 \text{ m/s}Answer
  4. D
    50 m/s50 \text{ m/s}

Answer

19.6 m/s19.6 \text{ m/s}
The correct answer applies the impulse-momentum theorem. First, the mass is found by dividing the weight by gravity (2,940 N/9.8 m/s2=300 kg2,940 \text{ N} / 9.8 \text{ m/s}^2 = 300 \text{ kg}). Then, the impulse is calculated as the braking force multiplied by time (588 N×10 s=5,880 N⋅s588 \text{ N} \times 10 \text{ s} = 5,880 \text{ N}\cdot\text{s}). Because impulse equals the change in momentum (m×Δvm \times \Delta v), dividing the impulse by the mass yields the sled's initial velocity (5,880 N⋅s/300 kg=19.6 m/s5,880 \text{ N}\cdot\text{s} / 300 \text{ kg} = 19.6 \text{ m/s}).

Step-by-Step Solution

1
Calculate the mass of the sled from its given weight.
m=2,940 N/9.8 m/s2=300 kgm = 2,940 \text{ N} / 9.8 \text{ m/s}^2 = 300 \text{ kg}
Momentum and kinematic equations require mass (in kg), not weight (in N).
2
Calculate the total impulse applied by the friction track.
Impulse=F×t=588 N×10 s=5,880 N⋅s\text{Impulse} = F \times t = 588 \text{ N} \times 10 \text{ s} = 5,880 \text{ N}\cdot\text{s}
Impulse is defined as the product of the average force and the time duration over which it acts.
3
Relate the impulse to the change in velocity.
5,880=300×Δv5,880 = 300 \times \Delta v, which solves to Δv=19.6 m/s\Delta v = 19.6 \text{ m/s}.
The impulse-momentum theorem states that impulse equals the change in momentum (Δp=m×Δv\Delta p = m \times \Delta v). Since the sled stops, the change in velocity equals its initial velocity.

Key Concept

Impulse and Momentum Conservation

Alternative Method

You can also solve this using Newton's Second Law and basic kinematics. First, find the deceleration: a=F/m=588/300=1.96 m/s2a = F / m = 588 / 300 = 1.96 \text{ m/s}^2. Then, use the kinematic equation vf=vi−a×tv_f = v_i - a \times t. Since vf=0v_f = 0, vi=a×t=1.96×10=19.6 m/sv_i = a \times t = 1.96 \times 10 = 19.6 \text{ m/s}.
Estimated Time:1m 30s
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Topics

Momentum and Collisions

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