Page 2 of a physics practice worksheet covering momentum calculations and conservation principles.

Here is the complete solution to all 11 problems on the worksheet, with detailed explanations for each.

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1. If a ball is rolling at a velocity of 1.5 m/sec and has a momentum of 10.0 kg·m/sec, what is the mass of the ball?

Solution:
Momentum (p) = mass (m) × velocity (v)

So,
> m = p / v

Plug in the values:
> m = 10.0 kg·m/s ÷ 1.5 m/s
> m = 6.67 kg

✔ Answer: 6.67 kg

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2. What is the velocity of an object that has a mass of 2.5 kilograms and a momentum of 1,000 kg·m/sec?

Solution:
Again, p = m × v → v = p / m

> v = 1000 kg·m/s ÷ 2.5 kg
> v = 400 m/s

✔ Answer: 400 m/s

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3. Tiger Woods hits 45.0-gram golf ball, giving it a speed of 75.0 m/sec. What momentum has Tiger given to the golf ball?

Solution:
First, convert mass to kilograms:
> 45.0 g = 0.045 kg

Then, p = m × v
> p = 0.045 kg × 75.0 m/s
> p = 3.375 kg·m/s

✔ Answer: 3.375 kg·m/s

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4. A 400-kilogram cannon fires a 10-kilogram cannonball at 20 m/sec. If the cannon is on wheels, at what velocity does it move backward? (This backward motion is called recoil velocity.)

Solution:
This is a conservation of momentum problem. Before firing, total momentum = 0 (everything at rest).

After firing:
> Momentum of cannonball + Momentum of cannon = 0
> (m_ball × v_ball) + (m_cannon × v_cannon) = 0

Let’s define forward as positive, so cannonball moves at +20 m/s, cannon recoils backward (negative direction).

> (10 kg)(20 m/s) + (400 kg)(v_cannon) = 0
> 200 + 400v_cannon = 0
> 400v_cannon = -200
> v_cannon = -0.5 m/s

The negative sign means it moves backward.

✔ Answer: 0.5 m/s backward

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5. "Big" Al stands on a skateboard at rest and throws a 0.5-kilogram rock at a velocity of 10.0 m/sec. "Big" Al moves back at 0.05 m/sec. What is the combined mass of "Big" Al and the skateboard?

Solution:
Conservation of momentum again. Initial momentum = 0.

Final momentum:
> Momentum of rock + Momentum of Al+skateboard = 0

Let M = combined mass of Al and skateboard.

> (0.5 kg)(10.0 m/s) + (M)(-0.05 m/s) = 0
> 5 - 0.05M = 0
> 0.05M = 5
> M = 5 / 0.05 = 100 kg

✔ Answer: 100 kg

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6. As the boat in which he is riding approaches a dock at 3.0 m/sec, Jasper stands up in the boat and jumps toward the dock. Jasper applies an average force of 800 newtons on the boat for 0.30 seconds as he jumps.

a. How much momentum does Jasper’s 80-kilogram body have as it lands on the dock?

b. What is Jasper’s speed on the dock?

Solution:

This is an impulse-momentum problem.

Impulse (J) = Force × time = change in momentum (Δp)

Jasper applies 800 N for 0.30 s on the boat — by Newton’s 3rd Law, the boat applies an equal and opposite force on Jasper.

So, impulse on Jasper = 800 N × 0.30 s = 240 N·s = 240 kg·m/s

This impulse changes Jasper’s momentum. Since he was initially moving with the boat at 3.0 m/s, his initial momentum was:

> p_initial = m × v = 80 kg × 3.0 m/s = 240 kg·m/s (forward)

The impulse from the boat adds 240 kg·m/s *in the direction toward the dock* (forward). So:

> p_final = p_initial + impulse = 240 + 240 = 480 kg·m/s

That’s the momentum as he lands on the dock.

a. Answer: 480 kg·m/s

Now, for part b:

> p = m × v → v = p / m = 480 kg·m/s ÷ 80 kg = 6.0 m/s

✔ b. Answer: 6.0 m/s

*(Note: This assumes the impulse is entirely in the direction of motion and no other forces act during the jump.)*

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7. Daryl the delivery guy gets out of his pizza delivery truck forgetting to set the parking brake. The 2,000 kilogram truck rolls down hill reaching a speed of 30 m/sec just before hitting a large oak tree. The vehicle stops 0.72 seconds after first making contact with the tree.

a. How much momentum does the truck have just before hitting the tree?

b. What is the average force applied by the tree?

Solution:

a. Momentum before impact:

> p = m × v = 2000 kg × 30 m/s = 60,000 kg·m/s

✔ a. Answer: 60,000 kg·m/s

b. Average force applied by the tree:

Use impulse-momentum theorem:
> Impulse = Δp = F_avg × t

Truck goes from 60,000 kg·m/s to 0 in 0.72 s.

> Δp = 0 - 60,000 = -60,000 kg·m/s (negative because force opposes motion)

> F_avg × 0.72 s = -60,000
> F_avg = -60,000 / 0.72 ≈ -83,333 N

Magnitude is 83,333 N; direction is opposite to motion.

✔ b. Answer: 83,333 N (or approximately 8.33 × 10⁴ N)

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8. Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of an average person is 60 kilograms and the mass of Earth is 5.98 × 10²⁴ kilograms:

a. What is the total momentum of the two billion people?

b. What is the effect of their action on Earth?

Solution:

a. Total momentum:

Number of people = 2 billion = 2 × 10⁹
Mass per person = 60 kg
Velocity = 7.0 m/s upward

> p_total = (2 × 10⁹) × 60 kg × 7.0 m/s
> p_total = 840 × 10⁹ kg·m/s = 8.4 × 10¹¹ kg·m/s upward

✔ a. Answer: 8.4 × 10¹¹ kg·m/s upward

b. Effect on Earth:

By conservation of momentum, Earth must gain an equal and opposite momentum downward.

So, Earth’s momentum change = -8.4 × 10¹¹ kg·m/s

Earth’s mass = 5.98 × 10²⁴ kg

> v_earth = p / m = (-8.4 × 10¹¹) / (5.98 × 10²⁴) ≈ -1.4 × 10⁻¹³ m/s

This is an incredibly tiny velocity — essentially undetectable.

✔ b. Answer: Earth recoils downward with a negligible speed of about 1.4 × 10⁻¹³ m/s.

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9. Tammy, a lifeguard, spots a swimmer struggling in the surf and jumps from her lifeguard chair to the sand beach. She makes contact with the sand at a speed of 6.00 m/sec leaving an indentation in the sand 0.10 meters deep.

a. If Tammy’s mass is 60. kilograms, what momentum as she first touches the sand?

b. What is the average force applied on Tammy by the sand beach?

Solution:

a. Momentum when she touches sand:

> p = m × v = 60 kg × 6.00 m/s = 360 kg·m/s

✔ a. Answer: 360 kg·m/s

b. Average force by sand:

We need to find the deceleration over 0.10 m.

Use kinematics:
> v_f² = v_i² + 2aΔx
> 0 = (6.00)² + 2a(0.10)
> 0 = 36 + 0.2a
> a = -36 / 0.2 = -180 m/s²

Then, F_avg = m × a = 60 kg × (-180 m/s²) = -10,800 N

Magnitude is 10,800 N.

Alternatively, using work-energy or impulse:
Time to stop can be found from Δx = (v_i + v_f)/2 × t → 0.10 = (6.00 + 0)/2 × t → t = 0.0333 s

Then F_avg = Δp / t = (0 - 360) / 0.0333 ≈ -10,800 N — same result.

✔ b. Answer: 10,800 N (upward, opposing motion)

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10. When a gun is fired, the shooter describes the sensation of the gun kicking. Explain this in terms of momentum conservation.

Solution:

When a gun is fired, the bullet gains forward momentum. To conserve total momentum (which was zero before firing), the gun must gain an equal amount of momentum in the opposite direction — backward. This backward motion is felt by the shooter as “kick” or “recoil.” The greater the bullet’s momentum, the greater the recoil.

✔ Answer: The gun recoils backward to conserve momentum — the bullet’s forward momentum is balanced by the gun’s backward momentum.

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11. What does it mean to say that momentum is conserved?

Solution:

Momentum is conserved when the total momentum of a system remains constant over time, provided no external net force acts on the system. In other words, in the absence of external forces, the sum of the momenta of all objects in a system before an interaction equals the sum after the interaction.

This is known as the Law of Conservation of Momentum.

✔ Answer: It means the total momentum of an isolated system remains constant — momentum is neither created nor destroyed, only transferred between objects within the system.

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✔ All problems solved with explanations. Let me know if you’d like diagrams or further clarification!