Fun with Gravity — Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. Define free fall and explain how it relates to Newton’s Second Law of Motion. Use an example to illustrate your answer.
Answer:
Free fall means an object moves under the influence of gravity alone, with no air resistance acting on it. In free fall near Earth’s surface, every object accelerates downward with the same constant acceleration called g (about 9.8 m/s²).
According to Newton’s Second Law (F = ma), the net force acting on an object equals its mass × acceleration. In free fall the only force is weight, which equals mg. Setting F = ma gives mg = ma, so the acceleration a = g, independent of the mass.
For example, when you toss a ball up, it slows, stops momentarily, and returns downward. While rising and falling (ignoring air resistance), the ball has acceleration g downward.
This shows mass cancels out in the equation, so in a vacuum a heavy and a light object fall together with the same acceleration. The key ideas to remember are free fall, only gravity, and acceleration = g.
Q2. How can we calculate the value of g from a weight measurement? Show the steps with an example and mention one limitation of this method.
Answer:
The relationship between weight (W) and mass (m) is W = mg, where g is acceleration due to gravity. Rearranging gives g = W / m.
Example: A 2 kg object has a measured weight of 19.6 N. Using the formula, g = 19.6 N / 2 kg = 9.8 m/s². This gives the local value of g.
Steps: (1) Measure the mass accurately using a balance. (2) Measure the weight with a spring balance or scale in newtons. (3) Divide weight by mass to find g.
Limitation: The method assumes the only force is gravity. Air buoyancy and errors in measuring weight (e.g., friction, calibration) can change the reading slightly. Also g varies a little with location (latitude and altitude), so the measured value is local rather than universal.
Q3. Why does a flat sheet of paper fall slower than a crumpled ball when dropped in air, but they fall together in a vacuum? Explain with the concept of air resistance and terminal velocity.
Answer:
In air, falling objects experience air resistance (drag), a force that opposes motion and depends on shape, area, and speed. A flat sheet of paper has a large surface area and catches more air, so drag is large and it falls slowly. A crumpled paper ball has a small area and streamlined shape, so it experiences less drag and falls faster.
In contrast, a vacuum has no air, so there is no drag force. Then the only force is gravity and both objects accelerate at g, falling together.
Terminal velocity occurs when drag equals weight and net acceleration becomes zero. A flat sheet reaches a small terminal velocity due to higher drag, while a compact ball reaches a higher terminal velocity.
Key points: air resistance causes different fall rates in air; in a vacuum all objects fall equally because only gravity acts.
Q4. Distinguish between mass and weight. If Amit’s weight at the North Pole is 700 N, what is his mass? Will his weight be the same at the equator? Explain briefly.
Answer:
Mass is the amount of matter in an object and is measured in kilograms (kg). It is a scalar and does not change with location.
Weight is the force of gravity on that mass, measured in newtons (N). It equals mg and can change if g changes with location.
To find Amit’s mass when weight is 700 N, use m = W / g. Taking g = 9.8 m/s², Amit’s mass = 700 / 9.8 ≈ 71.43 kg. His mass remains 71.43 kg everywhere.
His weight at the equator will be slightly less than at the pole because the effective g is a little smaller at the equator due to Earth’s rotation and the oblate shape (greater radius at equator). So weight changes slightly, but mass remains unchanged.
Q5. Describe how the Earth’s gravitational force affects different types of motion: free fall, projectile motion, and the motion of the Moon and planets.
Answer:
For free fall, Earth's gravity pulls objects straight down, causing a constant downward acceleration g (~9.8 m/s²). If air resistance is negligible, objects accelerate uniformly.
In projectile motion (e.g., a thrown ball), gravity acts vertically downward while the horizontal motion is unaffected (ignoring air). This combination produces a curved parabolic path. Vertical velocity changes due to g, while horizontal velocity remains nearly constant.
For orbital motion (Moon around Earth, planets around Sun), gravity acts as a centripetal force pulling the body toward the center. The object moves forward due to inertia but gravity continuously bends its path into an orbit. The object is essentially in continuous free fall around the central body.
In all these cases the direction of gravity is toward the center of the Earth (or central mass), and the strength of this pull determines the acceleration and the resulting motion.
High Complexity (Analytical & Scenario-Based)
Q6. Two spheres, A (mass 0.5 kg, radius large) and B (mass 2.0 kg, compact), are dropped from the same height in air. Explain, using physics concepts and equations qualitatively, which will hit the ground first and why. Mention the role of drag and terminal velocity.
Answer:
Gravity acts on both spheres producing downward force mg, so if there were no air, both would accelerate at g and land together. In air, there is an upward drag force that depends on speed, shape, and cross-sectional area: approximately F_drag ∝ C_d ρ A v² (where C_d is drag coefficient, ρ air density, A cross-section).
Sphere A has a large radius and therefore a large area A, so it experiences more drag at the same speed than compact sphere B. Sphere B’s smaller area and better shape mean less drag.
During fall, as speed increases, drag increases until drag = weight, producing terminal velocity v_t = sqrt( (2mg) / (C_d ρ A) ). Because B has larger mass and smaller area, its v_t is higher than A’s. Thus B reaches ground first because it can keep accelerating to a higher speed before drag balances weight.
The key idea: in air, drag and terminal velocity decide fall time; a heavy, compact object usually lands before a large-area, light object.
Q7. An astronaut drops a hammer and a feather from the same height on the Moon and then repeats the experiment on Earth (near sea level). Predict and explain the outcomes in both places. Calculate the fall time from 1.5 m for both bodies on Earth and on the Moon, using g_Earth = 9.8 m/s² and g_Moon = 1.62 m/s².
Answer:
On the Moon, there is no atmosphere, so no air resistance. Both the hammer and feather are in free fall and accelerate at the same rate (g_Moon = 1.62 m/s²), so they hit together.
On Earth, air resistance is significant. The feather experiences large drag relative to its weight and falls slowly, while the hammer is heavy and compact and falls quickly; the hammer reaches the ground much earlier.
Time to fall from height h in free fall: t = sqrt(2h / g). For h = 1.5 m:
On Earth: t_E = sqrt(2 × 1.5 / 9.8) ≈ sqrt(0.3061) ≈ 0.553 s.
On Moon: t_M = sqrt(2 × 1.5 / 1.62) ≈ sqrt(1.8519) ≈ 1.361 s.
So on the Moon both objects take ≈ 1.36 s and land together; on Earth, the hammer takes ≈ 0.55 s while a feather takes much longer due to air drag.
Q8. You want to determine the value of g by dropping a ball from different heights and measuring the fall time. Describe the experimental method, show the key formula, give a sample calculation, and list at least four precautions to reduce errors.
Answer:
Method: Measure several heights h (e.g., 1.0 m, 1.5 m, 2.0 m). For each height, release the ball from rest and measure the time t it takes to hit the ground using a stopwatch or electronic timer. Repeat each height multiple times and take the average time. Use the free-fall formula h = ½ g t², rearranged to g = 2h / t² to compute g for each trial and average the results.
Sample calculation: If h = 4.90 m and measured average t = 1.0 s, then g = 2 × 4.90 / (1.0)² = 9.8 m/s².
Precautions to reduce errors:
Use an electronic timer or light gate to avoid human reaction-time error.
Ensure the ball is released without initial push (use an electromagnet or clamp).
Minimize air currents and use a dense small ball to reduce air resistance.
Measure height accurately from the ball’s center to the impact point, and repeat trials to average random errors.
With careful technique, this simple method yields a good estimate of g, but air resistance and timing errors must be controlled.
Q9. A satellite in low Earth orbit seems to “float” around Earth without falling straight down. Using the idea of free fall and centripetal force, explain why the satellite does not crash into Earth. Include the relation between orbital speed and gravitational acceleration.
Answer:
A satellite in orbit is actually in continuous free fall toward Earth. However, it also has a tangential velocity (forward speed). Because of this sideways motion, while gravity pulls it inward, the satellite’s path curves such that it keeps missing the Earth — this balance creates an orbit.
Gravity provides the required centripetal force to bend the satellite’s path: mg_eff = m v² / r (for circular orbit), where v is orbital speed and r is distance from Earth's center. Using gravitational acceleration at that altitude, we set g(r) = v² / r.
If the satellite has the correct orbital speed (for Low Earth Orbit ~ *7.8 km/s...
