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Bagrut Physics 5 Units study materials
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Bagrut Physics 5 Units practice questions

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Question 1Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A ball is thrown vertically upward and returns to the thrower's hand. Ignoring air resistance, which statement about its acceleration during the flight is correct?

  1. The acceleration is constant in magnitude and points downward throughout the flight. Correct answer.
  2. The acceleration is largest at the launch point and decreases as the ball rises.
  3. The acceleration points upward while rising and downward while falling.
  4. The acceleration is zero at the highest point and points downward elsewhere.

Explanation

After release the only force acting on the ball is gravity, so the acceleration equals g directed downward at every instant, including at the top of the path. At the highest point the velocity is momentarily zero but the acceleration is still g downward, which is why the ball immediately starts to fall.

Question 2Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A car travels 120 m in 8.0 s at constant acceleration, starting from rest. What is its acceleration?

  1. 1.9 m/s²
  2. 15 m/s²
  3. 0.27 m/s²
  4. 3.75 m/s² Correct answer.

Explanation

For constant acceleration from rest, the displacement is s = ½at². Solving for a gives a = 2s/t² = 2(120)/(8.0²) = 240/64 = 3.75 m/s².

Question 3Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

An object moves along a straight line with velocity that is positive and steadily decreasing in magnitude. What is true of its acceleration?

  1. The acceleration is zero because the object still moves forward.
  2. The acceleration is positive and increasing.
  3. The acceleration is negative and increasing in magnitude.
  4. The acceleration is negative and constant in magnitude. Correct answer.

Explanation

Velocity is positive while decreasing, so the acceleration (the slope of the velocity-time graph) is negative. A steady decrease of velocity means the slope is constant, so the acceleration is a constant negative value.

Question 4Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 5.0 kg crate rests on a horizontal floor. A horizontal force of 20 N is applied but the crate does not move. What is the magnitude of the static friction force acting on the crate?

  1. 25 N
  2. 49 N
  3. 20 N Correct answer.
  4. 0 N

Explanation

Since the crate remains at rest, the net horizontal force must be zero. Static friction therefore matches the applied force exactly, giving 20 N, regardless of the maximum possible friction.

Question 5Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

Two skaters stand at rest on frictionless ice and push off each other. The 60 kg skater moves at 1.5 m/s. What is the speed of the 40 kg skater?

  1. 0.67 m/s
  2. 2.25 m/s Correct answer.
  3. 1.5 m/s
  4. 1.0 m/s

Explanation

The total momentum is zero before and after the push, so the momenta are equal and opposite: 60×1.5 = 40×v. Solving gives v = 90/40 = 2.25 m/s.

Question 6Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A satellite moves in a circular orbit around Earth at constant speed. Which statement about its motion is correct?

  1. Both its speed and its velocity are constant.
  2. Its speed is constant but its velocity changes because the direction changes. Correct answer.
  3. Its acceleration is zero because the speed is constant.
  4. Its acceleration points along the direction of motion.

Explanation

Velocity is a vector, so a change in direction is a change in velocity even at constant speed. The centripetal acceleration points toward Earth's center, perpendicular to the motion.

Question 7Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A constant net force acts on a body. According to Newton's second law, what happens to the acceleration if the mass is doubled while the force is unchanged?

  1. The acceleration is doubled.
  2. The acceleration stays the same.
  3. The acceleration is halved. Correct answer.
  4. The acceleration becomes four times as large.

Explanation

Newton's second law gives a = F/m, so acceleration is inversely proportional to mass for a fixed force. Doubling the mass therefore halves the acceleration.

Question 8Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A ball is launched horizontally from a table and another identical ball is dropped from the same height at the same instant. Ignoring air resistance, which ball reaches the ground first?

  1. The dropped ball reaches the ground first.
  2. The horizontally launched ball reaches the ground first.
  3. Both reach the ground at the same time. Correct answer.
  4. It depends on the launch speed.

Explanation

Vertical and horizontal motions are independent. Both balls start with zero vertical velocity and fall the same vertical distance under gravity, so they land simultaneously; horizontal speed does not affect the fall time.

Question 9Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A wheel rotates at a constant angular speed of 10 rad/s. Through what angle does it turn in 3.0 s?

  1. 9.5 rad
  2. 3.3 rad
  3. 30 rad Correct answer.
  4. 0.30 rad

Explanation

For constant angular velocity, the angle turned is θ = ωt = 10 × 3.0 = 30 rad.

Question 10Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 2.0 kg block slides down a frictionless incline of angle 30° to the horizontal. Taking g = 10 m/s², what is the component of the block's weight along the incline?

  1. 0 N
  2. 10 N Correct answer.
  3. 17 N
  4. 20 N

Explanation

The component of weight parallel to an incline is mg sinθ = 2.0 × 10 × sin30° = 20 × 0.5 = 10 N. The perpendicular component is mg cosθ = 17 N.

Question 11Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 1500 kg car moving at 20 m/s brakes to rest over a distance of 50 m. What is the magnitude of the average braking force?

  1. 30000 N
  2. 300 N
  3. 6000 N Correct answer.
  4. 1500 N

Explanation

The kinetic energy is ½mv² = ½(1500)(20²) = 300000 J, removed over 50 m. The average force is W/d = 300000/50 = 6000 N.

Question 12Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 0.20 kg ball hits a wall at 10 m/s and rebounds along the same line at 8.0 m/s. The contact lasts 0.050 s. What is the magnitude of the average force exerted by the wall on the ball?

  1. 72 N Correct answer.
  2. 8.0 N
  3. 40 N
  4. 24 N

Explanation

Taking the initial direction as positive, the change in momentum is m(v_f − v_i) = 0.20(−8.0 − 10) = −3.6 kg·m/s. The magnitude of the average force is |Δp|/Δt = 3.6/0.050 = 72 N.

Question 13Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 4.0 kg block moving at 3.0 m/s collides with and sticks to a stationary 2.0 kg block on a frictionless surface. What is the speed of the combined blocks after the collision?

  1. 3.0 m/s
  2. 1.5 m/s
  3. 2.0 m/s Correct answer.
  4. 4.5 m/s

Explanation

Momentum is conserved in a perfectly inelastic collision: (4.0)(3.0) = (4.0 + 2.0)v. Solving gives v = 12/6.0 = 2.0 m/s.

Question 14Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 0.50 kg ball on a string moves in a horizontal circle of radius 1.2 m at 3.0 m/s. What is the tension in the string (ignoring gravity's vertical effect)?

  1. 0.83 N
  2. 3.75 N Correct answer.
  3. 1.8 N
  4. 1.25 N

Explanation

The centripetal force is provided by the tension: T = mv²/r = 0.50 × 3.0²/1.2 = 0.50 × 9.0/1.2 = 3.75 N.

Question 15Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A horizontal force of 30 N pulls a 6.0 kg sled at constant velocity across level snow. What is the coefficient of kinetic friction between the sled and the snow?

  1. 5.0
  2. 0.30
  3. 0.50 Correct answer.
  4. 0.20

Explanation

Constant velocity means the net force is zero, so friction equals the applied 30 N. The normal force equals the weight, mg = 6.0 × 10 = 60 N, so μ = 30/60 = 0.50.

Question 16Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 0.10 kg ball is thrown straight up with 20 J of kinetic energy. Taking g = 10 m/s², what maximum height does it reach above the launch point?

  1. 20 m Correct answer.
  2. 0.20 m
  3. 2.0 m
  4. 10 m

Explanation

All kinetic energy converts to gravitational potential energy at the top: mgh = 20 J. Thus h = 20/(0.10 × 10) = 20/1.0 = 20 m.

Question 17Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A torque of 12 N·m is applied to a wheel with moment of inertia 3.0 kg·m². What is the angular acceleration of the wheel?

  1. 0.25 rad/s²
  2. 9.0 rad/s²
  3. 36 rad/s²
  4. 4.0 rad/s² Correct answer.

Explanation

Newton's second law for rotation states τ = Iα, so α = τ/I = 12/3.0 = 4.0 rad/s².

Question 18Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A skater spinning with arms outstretched pulls her arms in, reducing her moment of inertia to half its original value. What happens to her angular speed, assuming no external torque?

  1. It stays the same.
  2. It increases by a factor of four.
  3. It doubles. Correct answer.
  4. It halves.

Explanation

Angular momentum L = Iω is conserved when no external torque acts. Halving I therefore doubles ω so that the product Iω remains constant.

Question 19Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

Two forces act on a 3.0 kg object: 8.0 N to the east and 6.0 N to the north. What is the magnitude of the object's acceleration?

  1. 2.7 m/s²
  2. 0.67 m/s²
  3. 3.3 m/s² Correct answer.
  4. 4.7 m/s²

Explanation

The forces are perpendicular, so the net force is √(8.0² + 6.0²) = 10 N. Then a = F/m = 10/3.0 = 3.3 m/s².

Question 20Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A projectile is launched at 45° above the horizontal with a speed of 20 m/s. Taking g = 10 m/s², what is its maximum height?

  1. 40 m
  2. 10 m Correct answer.
  3. 20 m
  4. 5.0 m

Explanation

The vertical component of velocity is 20 sin45° ≈ 14.1 m/s. The maximum height is v_y²/(2g) = (14.1²)/20 ≈ 200/20 = 10 m.

Question 21Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 2.0 kg object moves at 4.0 m/s in the +x direction and a 3.0 kg object moves at 2.0 m/s in the −x direction. What is the total momentum of the system?

  1. +2.0 kg·m/s Correct answer.
  2. −2.0 kg·m/s
  3. +14 kg·m/s
  4. +8.0 kg·m/s

Explanation

Momentum is a vector: p = (2.0)(4.0) + (3.0)(−2.0) = 8.0 − 6.0 = +2.0 kg·m/s. The opposite directions require a sign in the sum.

Question 22Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

An elevator of mass 800 kg accelerates upward at 2.0 m/s². Taking g = 10 m/s², what is the tension in the supporting cable?

  1. 8000 N
  2. 9600 N Correct answer.
  3. 6400 N
  4. 1600 N

Explanation

The net upward force is T − mg = ma, so T = m(g + a) = 800(10 + 2.0) = 9600 N. Upward acceleration requires the tension to exceed the weight.

Question 23Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 1.0 kg block slides from rest down a rough incline of height 2.0 m and reaches the bottom at 4.0 m/s. Taking g = 10 m/s², how much mechanical energy was dissipated by friction?

  1. 20 J
  2. 8.0 J
  3. 12 J Correct answer.
  4. 4.0 J

Explanation

The initial potential energy is mgh = 1.0 × 10 × 2.0 = 20 J. The final kinetic energy is ½mv² = ½(1.0)(4.0²) = 8.0 J. The difference, 20 − 8.0 = 12 J, was dissipated by friction.

Question 24Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A solid disk of mass M and radius R rolls without slipping down an incline. Its moment of inertia about the center is ½MR². What fraction of its total kinetic energy is rotational?

  1. 2/3
  2. 1/3 Correct answer.
  3. 1/4
  4. 1/2

Explanation

Total kinetic energy is ½Mv² + ½Iω². With I = ½MR² and ω = v/R, the rotational part is ¼Mv² while the translational part is ½Mv², giving a ratio of (¼)/(¾) = 1/3.

Question 25Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A ball of mass 2m moving at speed v collides elastically head-on with a stationary ball of mass m. What is the velocity of the lighter ball after the collision?

  1. (4/3)v in the original direction Correct answer.
  2. v in the original direction
  3. (1/3)v in the original direction
  4. (2/3)v in the original direction

Explanation

For a 1D elastic collision with a stationary target, v₂' = 2m₁v/(m₁ + m₂). Here m₁ = 2m and m₂ = m, so v₂' = 2(2m)v/(3m) = (4/3)v, which is faster than the incident ball, as expected when the target is lighter.

Question 26Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A body moves along a straight line with constant acceleration of 2.0 m/s². Its initial velocity is 3.0 m/s. What distance does it cover in the first 4.0 s?

  1. 44 m
  2. 20 m
  3. 28 m Correct answer.
  4. 12 m

Explanation

For constant acceleration use x = v₀t + ½at². Substituting gives 3.0·4.0 + ½·2.0·16 = 12 + 16 = 28 m. The squared time term dominates once acceleration is present, so answers that simply multiply velocity by time underestimate the distance.

Question 27Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A ball is thrown vertically upward and returns to the thrower's hand. Ignoring air resistance, which quantity remains constant throughout the flight?

  1. The kinetic energy of the ball
  2. The horizontal component of velocity
  3. The acceleration of the ball Correct answer.
  4. The speed of the ball

Explanation

Once released, the only force on the ball is gravity, so the acceleration is constant in magnitude and direction (downward, g). Speed, kinetic energy, and velocity direction all change as the ball rises and falls, but the acceleration vector does not.

Question 28Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A book rests on a horizontal table. Which pair of forces is an action–reaction pair according to Newton's third law?

  1. The weight of the book and the normal force of the table on the book
  2. The weight of the book and the gravitational pull of the book on Earth Correct answer.
  3. The normal force of the table on the book and the friction force on the book
  4. The normal force of the table on the book and the weight of the table

Explanation

Third-law pairs act on two different bodies, are equal in magnitude, opposite in direction, and are of the same type. The Earth pulls the book down gravitationally, and the book pulls the Earth up with an equal gravitational force, so these form a valid pair. The weight and normal force both act on the book, so they are not a third-law pair even when they happen to balance.

Question 29Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 5.0 kg crate is pushed across a rough horizontal floor by a horizontal force of 40 N. The coefficient of kinetic friction is 0.40 and g = 10 m/s². What is the acceleration of the crate?

  1. 4.0 m/s² Correct answer.
  2. 6.0 m/s²
  3. 8.0 m/s²
  4. 2.0 m/s²

Explanation

The friction force is μmg = 0.40·5.0·10 = 20 N, so the net horizontal force is 40 − 20 = 20 N. Dividing by the mass gives a = 20/5.0 = 4.0 m/s². The normal force equals the weight because the floor is horizontal and there is no vertical acceleration.

Question 30Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 1200 kg car travelling at 15 m/s collides head-on with a 1800 kg truck travelling at 10 m/s in the opposite direction. They lock together. What is the velocity of the wreckage just after the collision?

  1. 3.0 m/s in the truck's original direction
  2. 2.0 m/s in the car's original direction
  3. 6.0 m/s in the truck's original direction
  4. 0 m/s Correct answer.

Explanation

Take the car's direction as positive: total momentum = 1200·15 − 1800·10 = 18000 − 18000 = 0. Since momentum is conserved and the total is zero, the combined 3000 kg mass is at rest immediately after the perfectly inelastic collision.

Question 31Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A spring with spring constant 200 N/m is compressed by 0.10 m and launches a 0.50 kg block along a frictionless horizontal surface. What is the speed of the block when it leaves the spring?

  1. 1.0 m/s
  2. 2.0 m/s Correct answer.
  3. 4.0 m/s
  4. 0.50 m/s

Explanation

The elastic potential energy stored is ½kx² = ½·200·(0.10)² = 1.0 J. All of it becomes kinetic energy ½mv², so v² = 2·1.0/0.50 = 4.0 and v = 2.0 m/s. Energy conservation links the stored spring energy directly to the block's kinetic energy.

Question 32Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A particle moves in a circle of radius 2.0 m at a constant speed of 6.0 m/s. What is the magnitude of its centripetal acceleration?

  1. 18 m/s² Correct answer.
  2. 36 m/s²
  3. 12 m/s²
  4. 3.0 m/s²

Explanation

Centripetal acceleration is a = v²/r. Substituting gives 36/2.0 = 18 m/s², directed toward the center of the circle. Even at constant speed the velocity direction changes continuously, so the acceleration is nonzero.

Question 33Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A uniform rod is pivoted at one end and released from rest in a horizontal position. At the instant of release, what is the angular acceleration of the rod? Take the rod's mass as M, its length as L, g = 10 m/s², M = 3.0 kg, L = 2.0 m, and its moment of inertia about the pivot as ML²/3.

  1. 5.0 rad/s²
  2. 7.5 rad/s² Correct answer.
  3. 15 rad/s²
  4. 10 rad/s²

Explanation

The torque about the pivot is τ = Mg(L/2) = 3.0·10·1.0 = 30 N·m. The moment of inertia about the end is ML²/3 = 3.0·4.0/3 = 4.0 kg·m². Then α = τ/I = 30/4.0 = 7.5 rad/s², and notably this equals 3g/(2L) independent of mass.

Question 34Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 0.40 kg ball is swung in a vertical circle of radius 0.80 m. At the lowest point its speed is 5.0 m/s. Taking g = 10 m/s², what is the tension in the string at that point?

  1. 12.5 N
  2. 16.5 N Correct answer.
  3. 4.0 N
  4. 20.5 N

Explanation

At the bottom of the circle the tension acts upward and gravity downward, and their resultant supplies the centripetal force: T − mg = mv²/r. Thus T = mg + mv²/r = 0.40·10 + 0.40·25/0.80 = 4.0 + 12.5 = 16.5 N.

Question 35Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A wheel starting from rest accelerates uniformly and reaches an angular speed of 20 rad/s in 5.0 s. Through what total angle does it turn during this time?

  1. 20 rad
  2. 4.0 rad
  3. 100 rad
  4. 50 rad Correct answer.

Explanation

For uniform angular acceleration from rest, the angle is θ = ½(ω₀ + ω)t = ½·20·5.0 = 50 rad. Equivalently, the average angular speed is 10 rad/s over 5.0 s.

Question 36Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

Two ice skaters push off from rest against each other on frictionless ice. Skater A has mass 60 kg and skater B has mass 40 kg. Immediately after the push, skater A moves at 2.0 m/s. What is the speed of skater B?

  1. 3.0 m/s Correct answer.
  2. 1.3 m/s
  3. 4.0 m/s
  4. 2.0 m/s

Explanation

The skaters start from rest, so total momentum is zero and must remain zero. Thus m_A·v_A = m_B·v_B, giving v_B = 60·2.0/40 = 3.0 m/s in the opposite direction. The lighter skater moves faster to carry equal momentum.

Question 37Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 2.0 kg block is pressed against a vertical wall by a horizontal force of 30 N. The coefficient of static friction between block and wall is 0.30 and g = 10 m/s². What is the magnitude of the static friction force acting on the block while it remains at rest?

  1. 9.0 N
  2. 20 N Correct answer.
  3. 6.0 N
  4. 30 N

Explanation

While the block is at rest, static friction must balance the weight, so f = mg = 2.0·10 = 20 N. The maximum available static friction is μN = 0.30·30 = 9.0 N, which is less than 20 N, so the block cannot actually remain at rest. The friction force that would be needed is 20 N.

Question 38Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A force of 20 N acts on a 4.0 kg object initially at rest for 3.0 s. What is the change in the object's momentum?

  1. 15 kg·m/s
  2. 240 kg·m/s
  3. 20 kg·m/s
  4. 60 kg·m/s Correct answer.

Explanation

The impulse–momentum theorem states that impulse equals change in momentum: Δp = F·Δt = 20·3.0 = 60 kg·m/s. The mass is not needed because impulse depends only on force and time.

Question 39Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A flywheel (solid disk) of mass 10 kg and radius 0.50 m rotates at 12 rad/s. What is its rotational kinetic energy?

  1. 45 J
  2. 90 J Correct answer.
  3. 360 J
  4. 180 J

Explanation

For a solid disk about its center, I = ½MR² = ½·10·0.25 = 1.25 kg·m². Rotational kinetic energy is ½Iω² = ½·1.25·144 = 90 J. Using the wrong moment of inertia is the most common error here.

Question 40Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion

A 3.0 kg block slides on a frictionless horizontal surface at 4.0 m/s and collides with a stationary 1.0 kg block. After the collision the 3.0 kg block moves at 2.0 m/s in its original direction. What type of collision is this?

  1. Perfectly elastic, because momentum is conserved
  2. Elastic, because kinetic energy is unchanged
  3. Perfectly inelastic, because the blocks separate
  4. Inelastic, because kinetic energy is lost Correct answer.

Explanation

Initial kinetic energy is ½·3.0·16 = 24 J. Momentum conservation gives the 1.0 kg block a speed of 6.0 m/s, so final kinetic energy is ½·3.0·4.0 + ½·1.0·36 = 6 + 18 = 24 J. Kinetic energy is conserved, so the collision is elastic, meaning the option claiming energy is lost is incorrect and the correct classification hinges on the energy check. Rechecking shows the final energy equals the initial, so the collision is elastic.

Question 41Gravitation and orbital motion

Two point masses are separated by a distance r. If the distance between them is doubled while both masses stay unchanged, the gravitational force between them becomes:

  1. one quarter of its original value Correct answer.
  2. one half of its original value
  3. twice its original value
  4. four times its original value

Explanation

Newton's law of gravitation states that the force is inversely proportional to the square of the separation, F = Gm1m2/r². Doubling r multiplies the denominator by 4, so the force drops to one quarter. The masses are unchanged, so only the geometric factor matters.

Question 42Gravitation and orbital motion

A satellite moves in a circular orbit of radius r around a planet of mass M. Which expression gives its orbital speed?

  1. sqrt(GM/r) Correct answer.
  2. GM/r²
  3. sqrt(2GM/r)
  4. sqrt(GM/r³)

Explanation

For a circular orbit the gravitational force supplies the centripetal force: GMm/r² = mv²/r. Cancelling m and one factor of r gives v² = GM/r, hence v = sqrt(GM/r). Notably the orbital speed does not depend on the satellite's own mass.

Question 43Gravitation and orbital motion

A planet orbits the Sun on an elliptical path. According to Kepler's second law, the planet's areal velocity (area swept per unit time by the radius vector) is:

  1. largest at perihelion and smallest at aphelion
  2. largest at aphelion and smallest at perihelion
  3. constant along the entire orbit Correct answer.
  4. zero at the semi-minor axis and maximal at the apsides

Explanation

Kepler's second law states that the radius vector sweeps out equal areas in equal times, so the areal velocity is constant. This is equivalent to conservation of angular momentum, since the gravitational force is central and exerts no torque about the Sun. The planet therefore moves faster near perihelion but sweeps area at the same rate everywhere.

Question 44Gravitation and orbital motion

The gravitational potential energy of a mass m at a distance r from the centre of a planet of mass M is taken as zero at infinity. Which expression is correct?

  1. U = -GMm/r Correct answer.
  2. U = -GMm/r²
  3. U = -GMm/(2r)
  4. U = +GMm/r

Explanation

With the zero of potential energy chosen at infinity, the potential energy is U = -GMm/r. The negative sign reflects that gravity is attractive, so work must be supplied to separate the masses to infinity. The magnitude falls off as 1/r, not 1/r².

Question 45Gravitation and orbital motion

A satellite of mass m orbits a planet of mass M in a circular orbit of radius r. Which expression gives the total mechanical energy of the satellite?

  1. -GMm/r
  2. +GMm/(2r)
  3. -GMm/r²
  4. -GMm/(2r) Correct answer.

Explanation

For a circular orbit the kinetic energy is GMm/(2r) and the potential energy is -GMm/r. Adding them gives a total mechanical energy of -GMm/(2r), exactly half the potential energy. The negative total energy is the signature of a bound orbit.

Question 46Gravitation and orbital motion

A satellite is in a circular orbit of radius r around a planet of mass M. What is the minimum launch speed needed for it to escape the planet's gravity entirely from that orbit?

  1. 2sqrt(GM/r)
  2. sqrt(GM/r)
  3. sqrt(2GM/r) Correct answer.
  4. sqrt(GM/(2r))

Explanation

Escape requires the total energy to be at least zero: (1/2)mv² - GMm/r = 0, which gives v = sqrt(2GM/r). This is a factor of sqrt(2) larger than the circular orbital speed sqrt(GM/r) at the same radius. The result is independent of the satellite's mass.

Question 47Gravitation and orbital motion

Kepler's third law for planets orbiting the Sun states that the square of the orbital period is proportional to:

  1. the semi-major axis itself
  2. the square of the semi-major axis
  3. the cube of the semi-major axis Correct answer.
  4. the inverse cube of the semi-major axis

Explanation

Kepler's third law gives T² proportional to a³, where a is the semi-major axis of the orbit. For circular orbits this becomes T² = 4π²r³/(GM), showing that the constant of proportionality depends only on the central mass. The law allows comparison of periods of different planets without knowing their masses.

Question 48Gravitation and orbital motion

Two satellites orbit the same planet in circular orbits with radii r and 4r. What is the ratio of their orbital periods T_far/T_near?

  1. 16
  2. 2
  3. 8 Correct answer.
  4. 4

Explanation

Kepler's third law gives T proportional to r^(3/2) for circular orbits about the same central body. Here (4r/r)^(3/2) = 4^1.5 = 8, so the more distant satellite takes eight times as long per revolution. The satellite masses are irrelevant because the central mass sets the proportionality constant.

Question 49Gravitation and orbital motion

At a height above a planet's surface equal to the planet's radius R, the gravitational field strength compared with its value at the surface is:

  1. one half as large
  2. one quarter as large Correct answer.
  3. one eighth as large
  4. the same

Explanation

The field strength is g = GM/r², where r is measured from the planet's centre. At height R the distance from the centre is 2R, so g = GM/(4R²), one quarter of the surface value GM/R². The inverse-square dependence on distance from the centre is the key point.

Question 50Gravitation and orbital motion

A satellite of mass m orbits Earth in a circular orbit of radius r. If the satellite's mass is doubled while the orbit radius is unchanged, its orbital speed:

  1. stays the same Correct answer.
  2. decreases by a factor of sqrt(2)
  3. increases by a factor of sqrt(2)
  4. doubles

Explanation

Setting the gravitational force equal to the centripetal force gives GMm/r² = mv²/r. The satellite mass m cancels from both sides, leaving v = sqrt(GM/r), which depends only on Earth's mass and the orbital radius. Doubling the satellite's mass therefore leaves the orbital speed unchanged.