Mass and Weight – Long Answer Questions

Medium Level (Application & Explanation)

Q1. Explain the difference between mass and weight with clear examples and units.

Answer:

Mass is the amount of matter in an object and is a scalar quantity. It is measured in kilograms (kg). Mass is constant for an object whether it is on Earth, on the Moon, or in space. For example, a stone of 5 kg has the same mass on Earth and on the Moon. Mass also gives a measure of inertia, which means how hard it is to change the object’s motion.
Weight is the force of gravity acting on an object and is a vector quantity. It is measured in newtons (N). Weight depends on the local gravitational acceleration (g) and is given by W = m × g. For example, the same 5 kg stone has weight W = 5 × 9.8 = 49 N on Earth, but on the Moon (g ≈ 1.63 m/s²) it weighs ≈ 8.15 N.
In short: mass does not change, weight changes with location and is a force.

Q2. How does a beam balance measure mass correctly on the Moon or on Earth, while a spring balance does not? Give practical reasoning.

Answer:

A beam balance compares the gravitational pull on an unknown mass with known standard masses. If both pans experience the same change in gravity, the relative effect cancels out so the balance still shows equality. That is why a beam balance gives the same mass reading on Earth, on the Moon, or at high altitude. The beam balance measures mass indirectly by comparing forces, and because those forces scale equally, the measured mass remains correct.
A spring balance works by measuring the stretch of a spring under weight; the stretch is proportional to force (weight), so its reading depends directly on g. On the Moon a 10 kg mass produces much less stretch and the spring balance shows a much smaller reading (weight), not the mass in kg unless calibrated for local gravity.
Practically, for measuring mass anywhere, a beam balance is reliable; for measuring weight or force, use a spring balance.

Q3. A 15 kg dumbbell is taken to the Moon. Calculate its weight on Earth and on the Moon, and explain why the weight differs while mass stays the same.

Answer:

The mass of the dumbbell is 15 kg and it remains 15 kg wherever it is.
On Earth, gravitational acceleration is g = 9.8 m/s². So weight W = m × g = 15 × 9.8 = 147 N.
On the Moon, gravity is about 1/6th of Earth’s. Using g_moon ≈ 1.63 m/s² or simply 1/6 of Earth’s value, weight on Moon ≈ (1/6) × 147 = 24.5 N (using 1/6) or 15 × 1.63 ≈ 24.45 N (using 1.63).
The reason the weight differs is that weight is a force due to gravity and depends on the local value of g. The dumbbell’s mass does not change because mass is the quantity of matter and is independent of location. So the same object can have the same mass but different weights on different celestial bodies.

Q4. Explain how mass is related to inertia. Give a practical classroom example showing the effect of different masses under the same force.

Answer:

Inertia is the tendency of an object to resist any change in its state of motion. Mass is the measure of inertia: the greater the mass, the greater the inertia, and the harder it is to change the object’s velocity.
Practical classroom example: Take a small toy car and a heavy metal block. Apply the same push (same force) to both. The toy car with small mass will accelerate noticeably, while the heavy block with large mass will accelerate very little. This follows Newton’s second law F = m × a; for the same force F, the acceleration a = F/m is smaller when m is larger.
Another example: Try to stop a rolling ball and then stop a rolling bowling ball. The bowling ball (larger mass) is harder to stop because of greater inertia.
Thus, mass determines how much an object resists changes in motion, and that is why heavy objects need more force to start or stop moving.

Q5. If you climb a mountain, does your mass or weight change? Explain the physics behind any change in weight with altitude.

Answer:

Your mass does not change when you climb a mountain; the amount of matter in your body stays the same. Mass is constant regardless of altitude.
Your weight can change slightly with altitude because gravitational acceleration (g) decreases with distance from the Earth's centre. The formula for weight is W = m × g. As altitude increases, the value of g becomes a little smaller, so weight decreases slightly.
For typical mountain heights (a few kilometres), the decrease in g and therefore weight is very small (a fraction of a percent), so you will not notice the change in day-to-day life. Only at very large distances (satellite altitude, Moon, other planets) does weight change become significant.
In summary: mass stays constant, weight decreases slightly with altitude because of weaker gravity.

High Complexity (Analytical & Scenario-Based)

Q6. A person of mass 60 kg stands on a weighing machine inside an elevator. Calculate the apparent weight when the elevator accelerates upward at 2 m/s², and when it accelerates downward at 2 m/s². Explain the concept of apparent weight in each case.

Answer:

Use g = 9.8 m/s². Real weight is mg = 60 × 9.8 = 588 N. Apparent weight equals the normal force N that the scale exerts and is found from Newton’s second law: N − mg = ma (take upward positive).
When elevator accelerates upward at a = +2 m/s²: N = m(g + a) = 60 × (9.8 + 2) = 60 × 11.8 = 708 N. The person feels heavier because the scale must push up harder to accelerate them upward.
When elevator accelerates downward at a = −2 m/s²: N = m(g − a) = 60 × (9.8 − 2) = 60 × 7.8 = 468 N. The person feels lighter because the scale pushes up with less force.
Apparent weight is the normal contact force that a surface provides to support you; it can be more or less than true weight depending on acceleration. In non-accelerating frames, apparent weight equals true weight. In accelerating frames, apparent weight changes because part of the gravitational requirement is used to accelerate the mass.

Q7. Astronauts in the International Space Station say they are “weightless.” Explain why they feel weightless even though Earth’s gravity is still acting on them. Include the role of free-fall and centripetal acceleration in your answer.

Answer:

In low Earth orbit the Earth’s gravity is still strong (only slightly less than on the surface), so objects are certainly not free of gravity. The station and astronauts are in continuous free-fall toward Earth, but they have a large horizontal velocity so they keep missing Earth — that is orbiting.
The station and everything inside it accelerate together at nearly the same rate under gravity. Because there is no contact force from the floor or walls to support the astronaut, a spring balance or scale reads zero, and the astronaut experiences weightlessness (no sensation of being pushed by a supporting surface).
The motion requires centripetal acceleration toward Earth to keep the station in orbit; gravity provides that centripetal force. Since the astronaut and station fall together, there is no relative contact force, so inhabitants feel floating or weightless even though gravity acts constantly.
Importantly, the astronaut’s mass remains unchanged; only the apparent weight (contact force) is effectively zero inside the freely falling spacecraft.

Q8. You are designing a 50 kg robot to work both on Earth and on the Moon. Calculate its weight on both places. Discuss at least three engineering considerations (traction, lifting actuation, stability) that must be adjusted because of the large difference in weight.

Answer:

First calculate weights: On Earth, W = m × g = 50 × 9.8 = 490 N. On the Moon (g ≈ 1.63 m/s² or 1/6 g), W ≈ 50 × 1.63 = 81.5 N (or 490/6 ≈ 81.67 N).
Engineering considerations:
- Traction and wheel design: Lower weight on the Moon means less normal force, so frictional force (µN) available for traction is reduced. Wheels or legs must be designed (wider treads, anchoring, or active gripping) to avoid slipping.
- Lifting and actuators: Because the robot is lighter on the Moon, lifting loads requires less force, so actuators can be smaller for vertical lifts; however, the robot might overshoot because of lower resistance, so motor control must be tuned for lower gravity and lower damping.
- Stability and center of mass: Lower weight reduces resisting moments from gravity, so the robot can tip more easily under lateral forces. Design must ensure a lower center of mass and possibly wider base or deployable outriggers on the Moon to maintain stability.
Additional considerations: braking systems, impact absorption (landings cause different dynamic responses), and energy budget (different work to climb slopes) must be adapted for the different weights even though mass is the same.

Q9. Compare readings of a beam balance and a spring balance if both are used to measure a mass inside a lift in three cases: (a) lift at rest, (b) lift accelerating upward, (c) lift in free-fall. Explain why readings differ or remain the same.

Answer:

(a) Lift at rest: Beam balance compares unknown mass with standard masses by comparing gravitational force; because gravity is normal, it balances and indicates the correct mass. Spring balance measures the weight (N) and will show mg. Both instruments work normally: beam balance shows mass correctly and spring balance shows weight.
(b) Lift accelerating upward (a upwards): The spring balance reading increases because it measures the normal force N = m(g + a). So weight reading is greater than mg. The beam balance st...