Momentum-Impulse | Summaries Acting | Docsity - Free Printable

Let's solve each problem step by step using the principles of momentum and impulse. The key formulas we'll use are:

- Impulse (J) = Force × Time = Δp (change in momentum)
- Momentum (p) = mass × velocity
- Average force = Impulse / time

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1. A 1,200-kilogram car traveling at 10 m/s hits a tree and is brought to rest in 0.10 second. What is the magnitude of the average force acting on the car?

We need to find the average force using:
$$
F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{m(v_f - v_i)}{\Delta t}
$$

Given:
- $ m = 1200 \, \text{kg} $
- $ v_i = 10 \, \text{m/s} $
- $ v_f = 0 \, \text{m/s} $
- $ \Delta t = 0.10 \, \text{s} $

$$
\Delta p = 1200 \times (0 - 10) = -12,000 \, \text{kg·m/s}
$$
$$
F_{\text{avg}} = \frac{-12,000}{0.10} = -120,000 \, \text{N}
$$

Magnitude: $ 1.2 \times 10^5 \, \text{N} $

✔ Answer: 4. $ 1.2 \times 10^5 \, \text{N} $

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2. A 50-kg student throws a 0.40-kg ball with a speed of 20 m/s. What was the magnitude of the impulse exerted on the ball?

Impulse = change in momentum of the ball.

Assuming the ball starts from rest:
$$
\Delta p = m \cdot \Delta v = 0.40 \, \text{kg} \times 20 \, \text{m/s} = 8.0 \, \text{kg·m/s} = 8.0 \, \text{N·s}
$$

✔ Answer: 1. $ 8.0 \, \text{N·s} $

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3. A 60-kg rollerskater exerts a 10-N force on a 30-kg rollerskater for 0.20 s. What is the magnitude of the impulse applied to the 30-kg skater?

Impulse = Force × Time
$$
J = F \cdot t = 10 \, \text{N} \times 0.20 \, \text{s} = 2.0 \, \text{N·s}
$$

Note: The mass of the skater doesn't matter here — impulse depends only on force and time.

✔ Answer: 2. $ 2.0 \, \text{N·s} $

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4. Two carts are pushed apart by a spring. Average force on 1-kg cart is 1 N. What is average force on 2-kg cart?

From Newton’s Third Law, the forces between two objects are equal and opposite.

So if the spring pushes the 1-kg cart with 1 N, it pushes the 2-kg cart with equal magnitude but opposite direction.

✔ Answer: 1. $ 1 \, \text{N} $

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5. What is the speed of a $ 1.0 \times 10^3 $ kg car that has a momentum of $ 2.0 \times 10^4 $ kg·m/s east?

Use:
$$
p = m \cdot v \Rightarrow v = \frac{p}{m}
$$

$$
v = \frac{2.0 \times 10^4}{1.0 \times 10^3} = 20 \, \text{m/s}
$$

So, $ 2.0 \times 10^1 \, \text{m/s} $

✔ Answer: 2. $ 2.0 \times 10^1 \, \text{m/s} $

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6. A 60-kg cyclist is projected at 20 m/s into a haystack and stops in 0.50 s. What is the average force exerted by the haystack?

Use:
$$
F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{m(v_f - v_i)}{\Delta t}
$$

- $ m = 60 \, \text{kg} $
- $ v_i = 20 \, \text{m/s} $
- $ v_f = 0 $
- $ \Delta t = 0.50 \, \text{s} $

$$
\Delta p = 60 \times (0 - 20) = -1200 \, \text{kg·m/s}
$$
$$
F_{\text{avg}} = \frac{-1200}{0.50} = -2400 \, \text{N}
$$

Magnitude: $ 2.4 \times 10^3 \, \text{N} $

✔ Answer: 4. $ 2.4 \times 10^3 \, \text{N} $

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7. A 70-kg hockey player is hit by a 0.1-kg puck moving west. The puck exerts a 50-N force toward the west on the player. Determine the magnitude of the force the player exerts on the puck.

Newton’s Third Law: Forces are equal and opposite.

So if the puck exerts 50 N west on the player, the player exerts 50 N east on the puck.

✔ Answer: $ 50 \, \text{N} $ (Not listed as an option — but the question says "determine", so likely expects 50 N)

But since it's multiple choice, let's check the options — none are given here. But based on physics, the answer is 50 N.

Wait — perhaps it's implied to be one of the choices? Let's assume the choices were missing. But the correct answer is 50 N.

✔ Answer: 50 N (This should be the correct choice.)

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8. Which situation produces the greatest change in momentum for a 1.0-kg cart?

Change in momentum $ \Delta p = m \cdot \Delta v $ or $ F \cdot t $

Let’s compute $ \Delta p $ for each:

1. From rest to 3.0 m/s → $ \Delta p = 1.0 \times 3.0 = 3.0 \, \text{kg·m/s} $
2. From 2.0 to 4.0 m/s → $ \Delta p = 1.0 \times (4.0 - 2.0) = 2.0 \, \text{kg·m/s} $
3. $ F = 5.0 \, \text{N}, t = 2.0 \, \text{s} $ → $ J = 5.0 \times 2.0 = 10.0 \, \text{N·s} = 10.0 \, \text{kg·m/s} $
4. $ F = 10.0 \, \text{N}, t = 0.5 \, \text{s} $ → $ J = 10.0 \times 0.5 = 5.0 \, \text{N·s} = 5.0 \, \text{kg·m/s} $

So the greatest is option 3: 10.0 kg·m/s

✔ Answer: 3. applying a net force of 5.0 N for 2.0 s

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9. A 0.149-kg baseball, initially moving at 15 m/s, is brought to rest in 0.040 s by a glove. What is the average force exerted on the ball?

Use:
$$
F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{m(v_f - v_i)}{\Delta t}
$$

- $ m = 0.149 \, \text{kg} $
- $ v_i = 15 \, \text{m/s} $
- $ v_f = 0 $
- $ \Delta t = 0.040 \, \text{s} $

$$
\Delta p = 0.149 \times (0 - 15) = -2.235 \, \text{kg·m/s}
$$
$$
F_{\text{avg}} = \frac{-2.235}{0.040} = -55.875 \, \text{N}
$$

Magnitude ≈ 56 N

✔ Answer: 4. $ 56 \, \text{N} $

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✔ Final Answers Summary:

| Question | Answer |
|--------|--------|
| 1 | 4. $ 1.2 \times 10^5 \, \text{N} $ |
| 2 | 1. $ 8.0 \, \text{N·s} $ |
| 3 | 2. $ 2.0 \, \text{N·s} $ |
| 4 | 1. $ 1 \, \text{N} $ |
| 5 | 2. $ 2.0 \times 10^1 \, \text{m/s} $ |
| 6 | 4. $ 2.4 \times 10^3 \, \text{N} $ |
| 7 | 50 N (Force is equal and opposite) |
| 8 | 3. applying a net force of 5.0 N for 2.0 s |
| 9 | 4. $ 56 \, \text{N} $ |

Let me know if you'd like explanations in simpler terms or diagrams!