Circular Motion & Gravitation — Practice · Grade 12 Science

Chapter: Forces, Rotation & Equilibrium

What this practice covers

By the end of Circular Motion & Gravitation you should be able to:

10 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: centripetal acceleration, centripetal force, period (T), Newton's law of universal gravitation, gravitational constant (G).

Problem 1

A car rounds a curve of radius 50 m at a speed of 10 m/s. Find its centripetal acceleration.
Answer and explanation

2 m/s²

a = v²/r = 10² ÷ 50 = 100 ÷ 50 = 2 m/s², directed toward the center of the curve.

Problem 2

A 2 kg ball moves in a horizontal circle of radius 4 m at 6 m/s. Find the centripetal force on it.
Answer and explanation

18 N

F = mv²/r = 2 × 6² ÷ 4 = 2 × 36 ÷ 4 = 72 ÷ 4 = 18 N.

Problem 3

An object moves in a circle at constant speed. In which direction does the net force on it point?
  1. Along the direction of motion (tangent to the circle)
  2. Outward, away from the center
  3. Toward the center of the circle
  4. There is no net force
Answer and explanation

Toward the center of the circle

Uniform circular motion requires a center-seeking (centripetal) net force. The outward feeling is an illusion of the accelerating frame.

Problem 4

If the distance between two masses is doubled, the gravitational force between them becomes…
  1. one-half as strong
  2. twice as strong
  3. unchanged
  4. one-quarter as strong
Answer and explanation

one-quarter as strong

Gravitation is an inverse-square law (F ∝ 1/r²). Doubling r multiplies the force by 1/2² = 1/4.

Problem 5

A satellite orbits Earth at constant speed. Explain why it is still accelerating, and name the force that provides its centripetal force.
Answer and explanation

Even at constant speed the satellite's direction is always changing, and any change in velocity is acceleration — here it is centripetal acceleration pointing toward Earth's center. The force supplying it is gravity: Earth's gravitational pull provides exactly the mv²/r needed to bend the satellite's path into a circle, so it 'falls around' the planet.

Problem 6

A 1,200 kg car rounds a flat curve of radius 80 m at 20 m/s. What centripetal force must friction supply?
Answer and explanation

6000 N

F = mv²/r = 1,200 × 20² ÷ 80 = 1,200 × 400 ÷ 80 = 1,200 × 5 = 6,000 N, directed toward the center of the curve.

Problem 7

At fixed mass and radius, how does doubling circular speed change required force? Show the reasoning, not just the conclusion.
Hints

Centripetal describes net inward force, not an additional mysterious interaction.

Answer and explanation

It quadruples force because F=mv²/r.

Problem 8

A satellite moves farther from the same spherical planet. How does gravitational field vary outside it?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

g=GM/r² decreases with distance from the center; doubling r gives one quarter the field.

Problem 9

Gravitational separation is measured between centersGravitational separation is measured between centerscenter-to-center rEarthF = GMm/r²; altitude h is not the same as r.For Earth radius R: r = R+h.
Original spherical-source separation model; Earth/satellite sizes and distances are not to scale.
Interpret the supplied visual. At altitude h=R, what fraction of surface g does this model predict?
Answer and explanation

r=2R, so g is one quarter of its surface value.

For ideal point masses or a spherical source outside its surface, Newtonian gravitation gives F=GMm/r² with center-to-center separation r. A satellite’s altitude h above a spherical Earth is not r; use r=R+h. Doubling center distance reduces force and gravitational acceleration to one quarter. The familiar near-surface g≈9.8 m/s² is an approximation over modest height differences, not the same value at every orbital distance.

r=2R, so g is one quarter of its surface value.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A 0.50 kg mass moves at 4.0 m/s in a 2.0 m horizontal circle. Find inward acceleration and required net force; name what else is needed to identify tension.
Answer and explanation

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N. Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N.

Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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