I85PREP
⌘ K
Pancyprian Physics study materials
Pancyprian Physics study resources

Pancyprian Physics practice questions

One question at a time. Build your Pancyprian Physics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A car travelling at 20 m/s brakes uniformly and stops in 5.0 s. What distance does it cover while braking?

  1. 40 m
  2. 50 m Correct answer.
  3. 100 m
  4. 10 m

Explanation

With uniform deceleration the average velocity is the mean of the initial and final velocities, (20 + 0)/2 = 10 m/s. Multiplying by the 5.0 s interval gives s = 10 × 5 = 50 m. Equivalently, s = ut + ½at² with a = −4 m/s² gives the same result.

Question 2Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

An object is dropped from rest and falls freely near the Earth's surface. Which statement about its motion is correct while air resistance is negligible?

  1. Its acceleration is zero because it started from rest.
  2. Its acceleration decreases as its speed grows.
  3. Its velocity increases by about 9.8 m/s every second. Correct answer.
  4. Its velocity stays constant while its acceleration grows.

Explanation

Free fall means the only force is weight, so the acceleration is constant at g ≈ 9.8 m/s² downward. A constant acceleration equal to 9.8 m/s² means the velocity changes by 9.8 m/s in each second. Starting from rest, the speed therefore grows steadily rather than levelling off.

Question 3Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 2.0 kg block rests on a horizontal table. What is the magnitude of the normal force the table exerts on the block?

  1. 2.0 N
  2. 0 N
  3. 9.8 N
  4. 19.6 N Correct answer.

Explanation

The block is in vertical equilibrium, so the normal force must balance the weight. The weight is mg = 2.0 × 9.8 = 19.6 N, and therefore the normal force is also 19.6 N upward. The normal force is a reaction to the surface being compressed, not to the block's mass directly.

Question 4Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A force of 12 N acts on a 3.0 kg body that is free to move on a frictionless horizontal surface. What is the body's acceleration?

  1. 36 m/s²
  2. 4.0 m/s² Correct answer.
  3. 0.25 m/s²
  4. 9.8 m/s²

Explanation

Newton's second law states that the net force equals mass times acceleration, so a = F/m. Substituting gives a = 12/3.0 = 4.0 m/s² in the direction of the applied force. Because the surface is frictionless, the applied force is the net horizontal force.

Question 5Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

Two ice skaters, initially at rest, push apart from each other. Which quantity is conserved for the system of the two skaters during the push?

  1. Total acceleration of the skaters
  2. Total speed of the skaters
  3. Total kinetic energy
  4. Total momentum Correct answer.

Explanation

The skaters exert equal and opposite forces on each other, so the net external horizontal force on the system is zero. Total momentum is therefore conserved and remains zero, meaning the skaters move apart with equal and opposite momenta. Kinetic energy is not conserved because chemical energy in the muscles is converted into motion.

Question 6Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A ball of mass 0.50 kg moving at 4.0 m/s is brought to rest in 0.20 s. What is the magnitude of the average force acting on the ball?

  1. 10 N Correct answer.
  2. 2.0 N
  3. 1.0 N
  4. 40 N

Explanation

The change in momentum is Δp = mΔv = 0.50 × 4.0 = 2.0 kg·m/s. The average force equals the rate of change of momentum, F = Δp/Δt = 2.0/0.20 = 10 N. This is the impulse–momentum theorem applied to the stopping process.

Question 7Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

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

  1. It points towards the Earth's centre along the radius. Correct answer.
  2. It is zero because the speed is constant.
  3. It points tangentially in the direction of motion.
  4. It points away from the Earth along the radius.

Explanation

In uniform circular motion the speed is constant but the direction of velocity changes continuously. The resulting centripetal acceleration always points towards the centre of the circle, here the Earth's centre, and has magnitude v²/r. A tangential acceleration would change the speed, which is not happening.

Question 8Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A crane lifts a 200 kg load vertically upward at constant speed through a height of 5.0 m. Taking g = 9.8 m/s², what is the work done by the crane's lifting force on the load?

  1. 0 J
  2. 1000 J
  3. 9800 J Correct answer.
  4. 200 J

Explanation

Constant speed means the lifting force equals the weight, mg = 200 × 9.8 = 1960 N. The work done by this force is Fd cos0° = 1960 × 5.0 ≈ 9800 J, which is also the gain in gravitational potential energy. The kinetic energy does not change because the speed is constant.

Question 9Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A projectile is launched at an angle to the horizontal over level ground with negligible air resistance. At the highest point of its path, which description of its velocity and acceleration is correct?

  1. Velocity is horizontal and acceleration is vertically downward. Correct answer.
  2. Velocity is vertically upward and acceleration is vertically downward.
  3. Velocity is zero and acceleration is zero.
  4. Velocity is horizontal and acceleration is zero.

Explanation

The horizontal component of velocity is constant throughout the flight because no horizontal force acts. At the highest point the vertical component has fallen to zero, so the velocity is purely horizontal. Gravity still acts, so the acceleration remains g downward at every instant of the flight.

Question 10Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A block slides down a rough inclined plane at constant velocity. Which statement about the forces on the block is correct?

  1. The component of weight along the plane equals the frictional force. Correct answer.
  2. Friction acts down the slope to help the motion.
  3. The normal force equals the weight of the block.
  4. The net force on the block points down the slope.

Explanation

Constant velocity means the net force along the plane must be zero. The weight component down the slope, mg sinθ, is therefore balanced by the kinetic friction acting up the slope, so the two are equal in magnitude. The normal force balances only the perpendicular weight component, mg cosθ.

Question 11Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A ball is thrown vertically upward from ground level with speed 15 m/s. Taking g = 9.8 m/s² and neglecting air resistance, what is the maximum height reached?

  1. 23 m
  2. 11 m Correct answer.
  3. 30 m
  4. 1.5 m

Explanation

At the highest point the speed is zero, so using v² = u² − 2gs gives 0 = 15² − 2 × 9.8 × s. Solving, s = 225/19.6 ≈ 11.5 m, which rounds to about 11 m. Energy conservation, ½mu² = mgh, gives the same height.

Question 12Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 1200 kg car rounds a flat horizontal curve of radius 50 m at 15 m/s. What is the minimum coefficient of static friction needed to keep the car on the curve?

  1. 0.30
  2. 0.46 Correct answer.
  3. 0.75
  4. 1.5

Explanation

The friction force provides the centripetal force, so μmg = mv²/r, giving μ = v²/(rg). Substituting, μ = 225/(50 × 9.8) ≈ 0.46. This is the minimum value; a larger coefficient would still keep the car on the curve.

Question 13Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 0.20 kg ball moving at 3.0 m/s collides head-on with a stationary 0.30 kg ball, and the two stick together. What is their common velocity after the collision?

  1. 1.5 m/s
  2. 1.2 m/s Correct answer.
  3. 2.0 m/s
  4. 3.0 m/s

Explanation

Momentum is conserved in the completely inelastic collision: m₁u₁ = (m₁ + m₂)v. Thus v = (0.20 × 3.0)/(0.50) = 0.60/0.50 = 1.2 m/s. Kinetic energy is not conserved, so this velocity must be found from momentum rather than energy.

Question 14Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 5.0 kg object is lifted from the ground to a shelf 2.0 m high and then allowed to fall back to the ground. Taking g = 9.8 m/s², what is its kinetic energy just before impact?

  1. 49 J
  2. 98 J Correct answer.
  3. 10 J
  4. 196 J

Explanation

By conservation of mechanical energy, the kinetic energy at impact equals the gravitational potential energy gained when lifted, mgh = 5.0 × 9.8 × 2.0 = 98 J. The path taken does not matter because gravity is a conservative force. Air resistance is neglected in this calculation.

Question 15Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

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

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

Explanation

Third-law pairs act on two different bodies and are of the same type. The Earth pulls the book downward with its weight, and the book pulls the Earth upward with an equal gravitational force; these form a genuine action–reaction pair. The normal force and the weight both act on the book, so they cannot be a third-law pair even when they are equal in magnitude.

Question 16Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A force of 20 N is applied to a 4.0 kg box on a rough horizontal floor, where the coefficient of kinetic friction is 0.25. Taking g = 9.8 m/s², what is the acceleration of the box?

  1. 0.26 m/s²
  2. 5.0 m/s²
  3. 2.6 m/s² Correct answer.
  4. 7.5 m/s²

Explanation

The friction force is μmg = 0.25 × 4.0 × 9.8 = 9.8 N, opposing the motion. The net force is 20 − 9.8 = 10.2 N, so a = 10.2/4.0 ≈ 2.6 m/s². Newton's second law must be applied to the net force, not the applied force alone.

Question 17Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A cyclist moves along a straight road with velocity that increases uniformly from 5.0 m/s to 15 m/s over 4.0 s. What is the displacement during this interval?

  1. 60 m
  2. 20 m
  3. 40 m Correct answer.
  4. 80 m

Explanation

The acceleration is (15 − 5.0)/4.0 = 2.5 m/s², and the displacement is s = ut + ½at² = 5.0 × 4.0 + ½ × 2.5 × 16 = 20 + 20 = 40 m. Alternatively, the average velocity is 10 m/s and 10 × 4.0 = 40 m.

Question 18Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A spring of spring constant 200 N/m is compressed by 0.10 m. What is the elastic potential energy stored in the spring?

  1. 2.0 J
  2. 20 J
  3. 1.0 J Correct answer.
  4. 10 J

Explanation

The energy stored in a stretched or compressed spring is ½kx². Substituting gives ½ × 200 × (0.10)² = ½ × 200 × 0.010 = 1.0 J. The energy depends on the square of the extension, so doubling the compression would quadruple the stored energy.

Question 19Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A stone is thrown horizontally from a cliff at 8.0 m/s. Taking g = 9.8 m/s² and neglecting air resistance, how far does it travel horizontally in the first 2.0 s?

  1. 8.0 m
  2. 20 m
  3. 39 m
  4. 16 m Correct answer.

Explanation

Horizontal and vertical motions are independent. With no horizontal force, the horizontal velocity stays at 8.0 m/s, so the horizontal distance is 8.0 × 2.0 = 16 m. The vertical fall affects only the vertical displacement, not the horizontal distance.

Question 20Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 1500 kg car moving at 20 m/s collides with a 1000 kg car at rest, and the two lock together. What fraction of the initial kinetic energy is lost in the collision?

  1. 40% Correct answer.
  2. 80%
  3. 60%
  4. 20%

Explanation

Momentum conservation gives a common velocity v = (1500 × 20)/2500 = 12 m/s. The initial kinetic energy is ½ × 1500 × 400 = 300 kJ, and the final is ½ × 2500 × 144 = 180 kJ. The loss is 120 kJ out of 300 kJ, which is 40%.

Question 21Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A particle moves in a circle of radius 0.50 m at a constant speed of 4.0 m/s. What is the magnitude of its centripetal acceleration?

  1. 16 m/s²
  2. 8.0 m/s²
  3. 2.0 m/s²
  4. 32 m/s² Correct answer.

Explanation

Centripetal acceleration is a = v²/r. Substituting gives a = (4.0)²/0.50 = 16/0.50 = 32 m/s². This acceleration points towards the centre of the circle and is caused by whatever force acts radially inward.

Question 22Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

Two blocks of masses 3.0 kg and 5.0 kg are connected by a light string over a frictionless pulley, with the 5.0 kg block hanging and the 3.0 kg block on a frictionless horizontal table. Taking g = 9.8 m/s², what is the acceleration of the system?

  1. 1.2 m/s²
  2. 6.1 m/s²
  3. 3.7 m/s² Correct answer.
  4. 9.8 m/s²

Explanation

The driving force is the weight of the hanging block, 5.0 × 9.8 = 49 N, and the total mass being accelerated is 8.0 kg. Thus a = 49/8.0 ≈ 6.1 m/s². The tension is found afterwards from either block using its own equation of motion.

Question 23Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A ball of mass m is dropped from rest from height h and rebounds to height h/4 after hitting the ground. What is the magnitude of the change in momentum of the ball during the bounce?

  1. (1/2)m√(2gh)
  2. 2m√(2gh)
  3. m√(2gh)
  4. (3/2)m√(2gh) Correct answer.

Explanation

Just before impact the speed is √(2gh) downward, and just after the bounce it is √(2g(h/4)) = ½√(2gh) upward. Taking upward as positive, the change in momentum is m(½√(2gh)) − m(−√(2gh)) = (3/2)m√(2gh). The signs matter because the velocity reverses direction.

Question 24Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A block of mass m slides from rest down a rough incline of height h and reaches the bottom with speed v. Which expression gives the work done against friction?

  1. mgh − ½mv² Correct answer.
  2. mgh + ½mv²
  3. ½mv² − mgh
  4. mgh

Explanation

The loss in gravitational potential energy is mgh, and the kinetic energy gained is ½mv². By the work–energy theorem, the difference between these quantities is the energy dissipated by friction, so the work done against friction is mgh − ½mv². This must be positive because friction always removes mechanical energy.

Question 25Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A car of mass 800 kg travels over the top of a hill of radius 40 m at 14 m/s. Taking g = 9.8 m/s², what is the magnitude of the normal force exerted by the road on the car at the top?

  1. 0 N
  2. 4.9 × 10³ N
  3. 2.9 × 10³ N Correct answer.
  4. 7.8 × 10³ N

Explanation

At the top of the hill the weight and normal force both act downward towards the centre, so mg − N = mv²/r. Thus N = m(g − v²/r) = 800 × (9.8 − 196/40) = 800 × (9.8 − 4.9) = 2.9 × 10³ N. The car loses contact with the road only if v²/r reaches g.

Question 26Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A runner completes one full lap of a circular track of radius 40 m. What is the magnitude of the runner's displacement for this lap?

  1. 0 m Correct answer.
  2. 80 m
  3. 251 m
  4. 40 m

Explanation

Displacement is the straight-line vector from the starting point to the finishing point. After one complete lap the runner returns to the starting point, so the displacement vector has zero length, even though the distance travelled equals the circumference, about 251 m.

Question 27Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A car travels 120 m in 8.0 s while moving at constant velocity. What is its velocity?

  1. 0.067 m/s
  2. 15 m/s Correct answer.
  3. 9.6 m/s
  4. 960 m/s

Explanation

For constant velocity, v = s/t = 120 m / 8.0 s = 15 m/s. The calculation is a direct division of the distance travelled by the time interval.

Question 28Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

An object starts from rest and accelerates uniformly at 3.0 m/s². How far does it travel in the first 4.0 s?

  1. 12 m
  2. 6.0 m
  3. 24 m Correct answer.
  4. 48 m

Explanation

Using s = ut + ½at² with u = 0 gives s = ½ × 3.0 × 4.0² = 24 m. The squared time term is essential: doubling the time quadruples the distance from rest.

Question 29Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 4.0 kg block rests on a horizontal floor. What is the magnitude of the normal force exerted by the floor on the block? Take g = 9.8 m/s².

  1. 9.8 N
  2. 39 N Correct answer.
  3. 4.0 N
  4. 0 N

Explanation

With no vertical acceleration on a horizontal surface, the normal force balances the weight: N = mg = 4.0 × 9.8 = 39.2 N, about 39 N. The normal force is not a fixed property of the surface but adjusts to the weight.

Question 30Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

Which quantity is a scalar?

  1. Momentum
  2. Weight
  3. Kinetic energy Correct answer.
  4. Acceleration

Explanation

Kinetic energy is defined by ½mv² and has magnitude only, with no direction, so it is a scalar. Momentum, acceleration and weight are all vectors because each has a direction as well as a magnitude.

Question 31Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

In an isolated system, the total momentum remains constant. This statement is a direct consequence of which law?

  1. Newton's second law applied to a single body
  2. Newton's first law
  3. Newton's third law Correct answer.
  4. Hooke's law

Explanation

Internal forces between interacting bodies occur in equal and opposite pairs acting for the same time, so their impulses cancel and total momentum is conserved. This pairing is exactly the content of Newton's third law.

Question 32Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A ball is thrown vertically upward. Neglecting air resistance, what is the ball's acceleration at the highest point of its path?

  1. 9.8 m/s² directed upward
  2. Zero
  3. It changes direction at that instant
  4. 9.8 m/s² directed downward Correct answer.

Explanation

Only gravity acts on the ball throughout the flight, so the acceleration is g directed downward at every instant, including the highest point. The velocity is momentarily zero there, but acceleration is the rate of change of velocity, not velocity itself.

Question 33Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A force of 12 N acts on a 3.0 kg object. What is the magnitude of the acceleration produced?

  1. 0.25 N/kg
  2. 0.25 m/s²
  3. 36 m/s²
  4. 4.0 m/s² Correct answer.

Explanation

Newton's second law gives a = F/m = 12 N / 3.0 kg = 4.0 m/s². Acceleration is directly proportional to the net force and inversely proportional to the mass.

Question 34Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 2.0 kg object moves in a straight line with velocity changing from 3.0 m/s to 9.0 m/s in 2.0 s. What is the magnitude of the net force on the object?

  1. 6.0 N Correct answer.
  2. 12 N
  3. 3.0 N
  4. 24 N

Explanation

The acceleration is (9.0 − 3.0)/2.0 = 3.0 m/s², and Newton's second law gives F = ma = 2.0 × 3.0 = 6.0 N. Alternatively, the change in momentum is 2.0 × 6.0 = 12 kg m/s, divided by 2.0 s gives the same force.

Question 35Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 60 kg person stands in a lift that accelerates upward at 2.0 m/s². Taking g = 9.8 m/s², what is the magnitude of the force the floor exerts on the person?

  1. 588 N
  2. 708 N Correct answer.
  3. 120 N
  4. 468 N

Explanation

Taking upward as positive, N − mg = ma, so N = m(g + a) = 60 × 11.8 = 708 N. The floor must both support the weight and provide the upward acceleration, so the normal force exceeds the weight.

Question 36Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 0.15 kg ball hits a wall horizontally at 8.0 m/s and rebounds horizontally at 6.0 m/s. What is the magnitude of the change in momentum of the ball?

  1. 0.90 kg m/s
  2. 1.2 kg m/s
  3. 0.30 kg m/s
  4. 2.1 kg m/s Correct answer.

Explanation

Taking the initial direction as positive, the change in momentum is m(v_final − v_initial) = 0.15 × (−6.0 − 8.0) = −2.1 kg m/s, so the magnitude is 2.1 kg m/s. The reversal of direction makes the change larger than the difference in speeds alone would suggest.

Question 37Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 0.40 kg ball moving at 5.0 m/s collides with a stationary 0.60 kg ball. After the collision the 0.40 kg ball moves at 2.0 m/s in the same direction. What is the speed of the 0.60 kg ball?

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

Explanation

Conservation of momentum gives 0.40 × 5.0 = 0.40 × 2.0 + 0.60 × v, so 2.0 = 0.80 + 0.60v and v = 2.0 m/s. The total momentum before and after the collision must be equal because no external horizontal force acts.

Question 38Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 1500 kg car accelerates from rest to 20 m/s. What is the kinetic energy gained by the car?

  1. 6.0 × 10⁵ J
  2. 1.5 × 10⁴ J
  3. 3.0 × 10⁴ J
  4. 3.0 × 10⁵ J Correct answer.

Explanation

The kinetic energy gained equals the final kinetic energy, ½mv² = ½ × 1500 × 20² = 3.0 × 10⁵ J. Because the speed is squared, doubling the final speed would quadruple the energy.

Question 39Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A constant horizontal force of 50 N pulls a 10 kg crate 4.0 m across a frictionless floor. What is the speed of the crate after 4.0 m if it started from rest?

  1. 2.0 m/s
  2. 20 m/s
  3. 6.3 m/s Correct answer.
  4. 40 m/s

Explanation

The work done is Fs = 50 × 4.0 = 200 J, which equals the kinetic energy gained. Then ½ × 10 × v² = 200 gives v² = 40 and v ≈ 6.3 m/s.

Question 40Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

Two forces act on a point: 6.0 N to the east and 8.0 N to the north. What is the magnitude of the resultant force?

  1. 14 N
  2. 48 N
  3. 2.0 N
  4. 10 N Correct answer.

Explanation

The two forces are perpendicular, so the resultant magnitude is √(6.0² + 8.0²) = √100 = 10 N. Vector addition is not simple arithmetic addition unless the forces are parallel.

Question 41Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 0.50 kg stone is whirled in a horizontal circle of radius 1.2 m at a constant speed of 3.0 m/s. What is the magnitude of the centripetal force?

  1. 1.25 N
  2. 0.63 N
  3. 7.5 N
  4. 3.75 N Correct answer.

Explanation

The centripetal force is F = mv²/r = 0.50 × 9.0 / 1.2 = 3.75 N. It points toward the centre of the circle and changes the direction of the velocity, not its magnitude.

Question 42Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A car travels at constant speed around a banked curve with no friction. Which force provides the centripetal acceleration?

  1. The forward driving force
  2. The horizontal component of the normal force Correct answer.
  3. The vertical component of the normal force
  4. The weight of the car

Explanation

On a frictionless banked curve the normal force is tilted from the vertical, and its horizontal component points toward the centre of the curve and supplies the centripetal force. The vertical component balances the weight.

Question 43Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 3.0 kg object moving at 4.0 m/s collides head-on with a 1.0 kg object moving at 2.0 m/s in the opposite direction. What is the total momentum of the system?

  1. 10 kg m/s in the direction of the 1.0 kg object
  2. 2.0 kg m/s in the direction of the 3.0 kg object
  3. 10 kg m/s in the direction of the 3.0 kg object Correct answer.
  4. 14 kg m/s in the direction of the 3.0 kg object

Explanation

Taking the 3.0 kg object's direction as positive, the total momentum is 3.0 × 4.0 + 1.0 × (−2.0) = 10 kg m/s. The opposite direction of the lighter object means its momentum is subtracted, not added.

Question 44Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A pump raises 200 kg of water per minute through a height of 10 m. Taking g = 9.8 m/s², what is the minimum output power of the pump?

  1. 19.6 kW
  2. 1960 W
  3. 327 W Correct answer.
  4. 54 W

Explanation

The work done each minute is mgh = 200 × 9.8 × 10 = 19 600 J in 60 s, so the power is 19 600/60 ≈ 327 W. Power is the rate of doing work, so the time interval must be included.

Question 45Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 2.0 kg block slides down a frictionless incline of height 3.0 m starting from rest. Taking g = 9.8 m/s², what is its speed at the bottom?

  1. 58 m/s
  2. 29 m/s
  3. 3.0 m/s
  4. 7.7 m/s Correct answer.

Explanation

Mechanical energy is conserved, so mgh = ½mv² and v = √(2gh) = √(2 × 9.8 × 3.0) ≈ 7.7 m/s. The mass cancels, so the answer does not depend on the 2.0 kg value.

Question 46Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

An object of mass 2.0 kg is attached to a spring of spring constant 50 N/m on a frictionless horizontal surface. The spring is stretched 0.20 m and released from rest. What is the maximum speed of the object?

  1. 5.0 m/s
  2. 0.50 m/s
  3. 2.0 m/s
  4. 1.0 m/s Correct answer.

Explanation

The stored elastic energy ½kx² = ½ × 50 × 0.04 = 1.0 J converts entirely into kinetic energy at the equilibrium position. Then ½ × 2.0 × v² = 1.0 gives v = 1.0 m/s.

Question 47Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A ball of mass 0.20 kg falls from rest through 1.8 m and bounces back to a height of 0.80 m. Taking g = 9.8 m/s², what is the magnitude of the impulse delivered by the ground?

  1. 1.3 N s
  2. 0.80 N s
  3. 3.3 N s
  4. 2.0 N s Correct answer.

Explanation

The impact speed is √(2 × 9.8 × 1.8) ≈ 5.94 m/s downward and the rebound speed is √(2 × 9.8 × 0.80) ≈ 3.96 m/s upward. The change in momentum has magnitude 0.20 × (5.94 + 3.96) ≈ 2.0 N s, and the impulse from the ground equals this change plus the small gravity impulse during contact, so 2.0 N s is the required answer.

Question 48Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

Two masses, 4.0 kg and 6.0 kg, hang from a light string over a frictionless pulley. Taking g = 9.8 m/s², what is the acceleration of the system?

  1. 1.96 m/s² Correct answer.
  2. 3.92 m/s²
  3. 9.8 m/s²
  4. 0.98 m/s²

Explanation

The net driving force is (6.0 − 4.0) × 9.8 = 19.6 N and the total mass accelerated is 10.0 kg, so a = 19.6/10.0 = 1.96 m/s². This is g multiplied by the mass difference divided by the mass sum.

Question 49Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A block is pushed up a rough incline and then slides back down. The coefficient of kinetic friction is 0.30 and the incline makes 30° with the horizontal. Taking g = 9.8 m/s², what is the magnitude of the acceleration of the block while it slides up the incline?

  1. 9.8 m/s²
  2. 7.4 m/s² Correct answer.
  3. 4.9 m/s²
  4. 2.4 m/s²

Explanation

While sliding up, both the component of weight along the slope and friction oppose the motion, so a = g(sinθ + μcosθ) = 9.8(0.50 + 0.30 × 0.866) ≈ 7.4 m/s². Friction acts down the slope during the upward slide.

Question 50Mechanics (kinematics, dynamics, Newton's laws, momentum, energy, circular motion)

A 1200 kg car travels at 20 m/s over the top of a hill whose radius of curvature is 50 m. Taking g = 9.8 m/s², at what minimum speed would the car lose contact with the road at the top?

  1. 9.8 m/s
  2. 22 m/s Correct answer.
  3. 15 m/s
  4. 49 m/s

Explanation

At the top, mg − N = mv²/r. Contact is lost when N becomes zero, giving v = √(gr) = √(9.8 × 50) ≈ 22 m/s. The mass of the car cancels out of the condition.