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NY Regents Physics study materials
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NY Regents Physics practice questions

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Question 1Mechanics

A car travels 120 meters in 15 seconds at a constant speed. What is the car's speed?

  1. 105 m/s
  2. 8.0 m/s Correct answer.
  3. 0.13 m/s
  4. 1800 m/s

Explanation

Average speed equals total distance divided by total time, so 120 m ÷ 15 s = 8.0 m/s. This is a scalar quantity describing how fast the car moves regardless of direction.

Question 2Mechanics

Which of the following is a vector quantity?

  1. Distance
  2. Time
  3. Displacement Correct answer.
  4. Speed

Explanation

A vector has both magnitude and direction, and displacement qualifies because it describes a change in position with a specified direction. Speed, distance, and time are scalars with magnitude only.

Question 3Mechanics

An object accelerates uniformly from rest to 20 m/s in 4.0 seconds. What is its acceleration?

  1. 0.20 m/s²
  2. 80 m/s²
  3. 5.0 m/s² Correct answer.
  4. 16 m/s²

Explanation

Acceleration is the change in velocity divided by the time interval, so (20 m/s − 0 m/s) ÷ 4.0 s = 5.0 m/s². Uniform acceleration means this rate stays constant throughout the interval.

Question 4Mechanics

According to Newton's first law, an object in motion continues in a straight line at constant speed unless acted on by a net external force. This property is best described as

  1. Impulse
  2. Momentum
  3. Weight
  4. Inertia Correct answer.

Explanation

Inertia is the tendency of an object to resist changes in its state of motion, which is exactly what Newton's first law describes. Mass is the quantitative measure of inertia.

Question 5Mechanics

A 10-newton force acts on a 2.0-kilogram mass. What is the resulting acceleration?

  1. 5.0 m/s² Correct answer.
  2. 20 m/s²
  3. 0.20 m/s²
  4. 12 m/s²

Explanation

Newton's second law states that net force equals mass times acceleration, so acceleration equals force divided by mass: 10 N ÷ 2.0 kg = 5.0 m/s². The direction of acceleration matches the direction of the net force.

Question 6Mechanics

Two forces act on an object: 6 N to the right and 4 N to the left. What is the magnitude and direction of the net force?

  1. 2 N to the left
  2. 2 N to the right Correct answer.
  3. 10 N to the left
  4. 10 N to the right

Explanation

Forces in opposite directions subtract, giving a net force of 6 N − 4 N = 2 N. The net force points in the direction of the larger force, which is to the right.

Question 7Mechanics

A 5.0-kilogram object rests on a horizontal surface where the coefficient of static friction is 0.40. Using g = 9.8 m/s², what is the maximum static friction force?

  1. 20 N Correct answer.
  2. 2.0 N
  3. 49 N
  4. 0.20 N

Explanation

Maximum static friction equals the coefficient of static friction times the normal force, and on a horizontal surface the normal force equals weight: 0.40 × (5.0 kg × 9.8 m/s²) = 0.40 × 49 N ≈ 20 N.

Question 8Mechanics

A 1500-kilogram car moving at 20 m/s comes to rest in 5.0 seconds. What is the magnitude of the average braking force?

  1. 300 N
  2. 7500 N
  3. 30 N
  4. 6000 N Correct answer.

Explanation

First find acceleration: (0 − 20 m/s) ÷ 5.0 s = −4.0 m/s². Then apply Newton's second law: force magnitude = 1500 kg × 4.0 m/s² = 6000 N.

Question 9Mechanics

A ball is thrown straight up with an initial velocity of 15 m/s. Using g = 9.8 m/s² and ignoring air resistance, how long does it take to reach its highest point?

  1. 1.5 s Correct answer.
  2. 2.0 s
  3. 0.65 s
  4. 4.9 s

Explanation

At the highest point the vertical velocity is zero, so using v = v₀ − gt gives 0 = 15 − 9.8t, and t ≈ 1.5 s. Gravity decelerates the ball at 9.8 m/s² throughout the upward trip.

Question 10Mechanics

A projectile is launched horizontally from a height and follows a curved path. Which statement about its motion is correct?

  1. The horizontal velocity changes while the vertical velocity stays constant.
  2. The horizontal velocity stays constant while the vertical velocity changes. Correct answer.
  3. Both horizontal and vertical velocities change at the same rate.
  4. Both horizontal and vertical velocities remain constant.

Explanation

Ignoring air resistance, no horizontal force acts on the projectile, so horizontal velocity is constant. Gravity acts vertically, so the vertical velocity changes at 9.8 m/s² downward.

Question 11Mechanics

A 2.0-kilogram object moving at 3.0 m/s has what momentum?

  1. 18 kg·m/s
  2. 1.5 kg·m/s
  3. 0.67 kg·m/s
  4. 6.0 kg·m/s Correct answer.

Explanation

Momentum equals mass times velocity: 2.0 kg × 3.0 m/s = 6.0 kg·m/s. Momentum is a vector pointing in the direction of the velocity.

Question 12Mechanics

A 4.0-kilogram cart moving at 3.0 m/s collides with and sticks to a stationary 2.0-kilogram cart. What is their combined velocity after the collision?

  1. 3.0 m/s
  2. 6.0 m/s
  3. 2.0 m/s Correct answer.
  4. 1.0 m/s

Explanation

In a perfectly inelastic collision, momentum is conserved: (4.0 kg)(3.0 m/s) + 0 = (6.0 kg)v, so v = 12 ÷ 6.0 = 2.0 m/s. The combined mass after sticking is 6.0 kg.

Question 13Mechanics

A 0.50-kilogram ball is dropped from rest and reaches a speed of 10 m/s just before hitting the ground. What is its kinetic energy at that moment?

  1. 2.5 J
  2. 5.0 J
  3. 50 J
  4. 25 J Correct answer.

Explanation

Kinetic energy equals one-half mass times velocity squared: ½ × 0.50 kg × (10 m/s)² = 0.25 × 100 = 25 J. Energy depends on the square of the speed.

Question 14Mechanics

A 2.0-kilogram book is lifted 1.5 meters above a table. Using g = 9.8 m/s², what is its increase in gravitational potential energy?

  1. 3.0 J
  2. 0.31 J
  3. 13 J
  4. 29 J Correct answer.

Explanation

Gravitational potential energy change equals mass times g times height: 2.0 kg × 9.8 m/s² × 1.5 m ≈ 29 J. This energy is stored due to the book's raised position.

Question 15Mechanics

A machine does 400 joules of work in 8.0 seconds. What is its power output?

  1. 3200 W
  2. 408 W
  3. 0.020 W
  4. 50 W Correct answer.

Explanation

Power is work divided by time: 400 J ÷ 8.0 s = 50 W. Power measures the rate at which energy is transferred or work is done.

Question 16Mechanics

A box is pushed 5.0 meters across a floor by a 20-newton horizontal force. How much work is done by this force?

  1. 25 J
  2. 4.0 J
  3. 100 J Correct answer.
  4. 0 J

Explanation

Work equals force times displacement when the force is parallel to the displacement: 20 N × 5.0 m = 100 J. Work measures energy transferred to the box.

Question 17Mechanics

A 0.25-kilogram ball moving at 8.0 m/s strikes a wall and rebounds straight back at 6.0 m/s. What is the magnitude of the ball's change in momentum?

  1. 0.50 kg·m/s
  2. 3.5 kg·m/s Correct answer.
  3. 2.0 kg·m/s
  4. 14 kg·m/s

Explanation

Taking the initial direction as positive, initial momentum is 0.25 × 8.0 = 2.0 kg·m/s and final momentum is 0.25 × (−6.0) = −1.5 kg·m/s. The change is −1.5 − 2.0 = −3.5 kg·m/s, with magnitude 3.5 kg·m/s.

Question 18Mechanics

A 3.0-kilogram block slides down a frictionless incline that is 2.0 meters high. Using g = 9.8 m/s², what is its speed at the bottom?

  1. 39 m/s
  2. 6.3 m/s Correct answer.
  3. 19.6 m/s
  4. 4.4 m/s

Explanation

By conservation of energy, mgh = ½mv², so v = √(2gh) = √(2 × 9.8 × 2.0) = √39.2 ≈ 6.3 m/s. Mass cancels, so the speed is independent of the block's mass.

Question 19Mechanics

A 1200-kilogram car rounds a flat curve of radius 50 meters at 15 m/s. What is the magnitude of the centripetal force required?

  1. 5400 N Correct answer.
  2. 240 N
  3. 18000 N
  4. 360 N

Explanation

Centripetal force equals mass times velocity squared divided by radius: (1200 kg × (15 m/s)²) ÷ 50 m = (1200 × 225) ÷ 50 = 5400 N. This force points toward the center of the circular path.

Question 20Mechanics

Two students push a 30-kilogram sled from rest with a combined net force of 60 N over a distance of 4.0 meters on frictionless ice. What is the sled's final kinetic energy?

  1. 0 J
  2. 1800 J
  3. 15 J
  4. 240 J Correct answer.

Explanation

The work done by the net force equals the change in kinetic energy: W = Fd = 60 N × 4.0 m = 240 J. Starting from rest, this work becomes the sled's final kinetic energy.

Question 21Mechanics

A 1.5-kilogram object hangs at rest from a single vertical rope. Using g = 9.8 m/s², what is the tension in the rope?

  1. 6.5 N
  2. 0 N
  3. 1.5 N
  4. 15 N Correct answer.

Explanation

Because the object is at rest, the net force is zero, so the upward tension equals the downward weight: T = mg = 1.5 kg × 9.8 m/s² ≈ 15 N.

Question 22Mechanics

A student measures the period of a pendulum as 2.0 seconds. What is the frequency of the pendulum?

  1. 0.25 Hz
  2. 2.0 Hz
  3. 4.0 Hz
  4. 0.50 Hz Correct answer.

Explanation

Frequency is the reciprocal of period: f = 1/T = 1 ÷ 2.0 s = 0.50 Hz. Frequency counts the number of complete oscillations per second.

Question 23Mechanics

A 0.20-kilogram arrow is shot from a bow with a speed of 30 m/s. What is the arrow's kinetic energy?

  1. 180 J
  2. 3.0 J
  3. 6.0 J
  4. 90 J Correct answer.

Explanation

Kinetic energy equals ½mv² = ½ × 0.20 kg × (30 m/s)² = 0.10 × 900 = 90 J. The energy scales with the square of the speed.

Question 24Mechanics

A 5.0-newton force acts on a 2.0-kilogram object for 3.0 seconds. What is the magnitude of the impulse delivered?

  1. 30 N·s
  2. 1.7 N·s
  3. 6.0 N·s
  4. 15 N·s Correct answer.

Explanation

Impulse equals force times the time interval: 5.0 N × 3.0 s = 15 N·s. Impulse also equals the change in momentum of the object.

Question 25Mechanics

An object moves along a straight line, and its velocity-time graph is a horizontal line above the time axis. Which description of the motion is correct?

  1. The object's acceleration increases steadily.
  2. The object accelerates at a constant nonzero rate.
  3. The object is at rest the entire time.
  4. The object moves with constant velocity and zero acceleration. Correct answer.

Explanation

On a velocity-time graph, the slope represents acceleration. A horizontal line has zero slope, so acceleration is zero and velocity stays constant.

Question 26Mechanics

A car travels 120 meters in 8.0 seconds. What is its average speed?

  1. 8.0 m/s
  2. 12 m/s
  3. 15 m/s Correct answer.
  4. 10 m/s

Explanation

Average speed is the total distance traveled divided by the total time taken. Here, 120 m divided by 8.0 s gives 15 m/s.

Question 27Mechanics

An object is in equilibrium. Which statement must be true about the forces acting on it?

  1. The forces are all equal in magnitude.
  2. The object is at rest.
  3. The net force is zero. Correct answer.
  4. The object is moving at constant velocity.

Explanation

Equilibrium means the vector sum of all forces acting on the object is zero. This can occur whether the object is at rest or moving with constant velocity, and the individual forces need not be equal.

Question 28Mechanics

A ball is thrown straight up with an initial speed of 20 m/s. Ignoring air resistance and using g = 9.8 m/s², what is the acceleration of the ball at the highest point of its path?

  1. 0 m/s²
  2. 9.8 m/s² downward Correct answer.
  3. 20 m/s² downward
  4. 9.8 m/s² upward

Explanation

At the highest point, the ball momentarily stops, but gravity still acts on it. The acceleration due to gravity is always 9.8 m/s² downward near Earth's surface, regardless of the ball's velocity.

Question 29Mechanics

A 10-newton force and a 6.0-newton force act concurrently on an object. Which pair of forces could be the equilibrant that keeps the object at rest?

  1. A 6.0 N force opposite to the 10 N force
  2. A 10 N force opposite to the 6.0 N force
  3. A 4.0 N force opposite to the 6.0 N force Correct answer.
  4. A 16 N force opposite to the 10 N force

Explanation

For equilibrium, the vector sum of all forces must be zero. The 10 N and 6.0 N forces in the same direction have a resultant of 16 N, so the equilibrant must be 16 N in the opposite direction. The option stating a 4.0 N force opposite to the 6.0 N force would combine with the 10 N force to give a net of 8.0 N, not zero, so it is incorrect.

Question 30Mechanics

A 2.0-kilogram object is pushed with a constant net force of 8.0 newtons. What is the acceleration of the object?

  1. 0.25 m/s²
  2. 16 m/s²
  3. 4.0 m/s² Correct answer.
  4. 8.0 m/s²

Explanation

According to Newton's second law, acceleration equals net force divided by mass. Here, 8.0 N divided by 2.0 kg gives 4.0 m/s².

Question 31Mechanics

A 0.50-kilogram cart moving at 2.0 m/s collides with a stationary 1.5-kilogram cart and they stick together. What is the speed of the combined carts immediately after the collision?

  1. 1.0 m/s
  2. 0.67 m/s
  3. 2.0 m/s
  4. 0.50 m/s Correct answer.

Explanation

In a perfectly inelastic collision, momentum is conserved. The initial momentum is (0.50 kg)(2.0 m/s) = 1.0 kg·m/s, and the total mass after collision is 2.0 kg, so the final speed is 1.0 kg·m/s ÷ 2.0 kg = 0.50 m/s.

Question 32Mechanics

A 1.0-kilogram object is lifted vertically at a constant speed of 2.0 m/s for 3.0 seconds. Using g = 9.8 m/s², how much work is done by the lifting force?

  1. 6.0 J
  2. 59 J Correct answer.
  3. 29 J
  4. 20 J

Explanation

The lifting force must equal the weight, mg = (1.0 kg)(9.8 m/s²) = 9.8 N, because the object moves at constant speed. The distance lifted is (2.0 m/s)(3.0 s) = 6.0 m, so the work done is (9.8 N)(6.0 m) ≈ 59 J.

Question 33Mechanics

A 0.20-kilogram ball is thrown vertically upward with 10 joules of kinetic energy. Ignoring air resistance and using g = 9.8 m/s², what maximum height does the ball reach?

  1. 2.0 m
  2. 0.51 m
  3. 10 m
  4. 5.1 m Correct answer.

Explanation

At the maximum height, all kinetic energy is converted to gravitational potential energy. Setting mgh = 10 J gives h = 10 J / [(0.20 kg)(9.8 m/s²)] ≈ 5.1 m.

Question 34Mechanics

A 1500-kilogram car accelerates uniformly from rest to 20 m/s in 10 seconds. What is the average power delivered by the net force during this interval?

  1. 15,000 W
  2. 30,000 W Correct answer.
  3. 3,000 W
  4. 300,000 W

Explanation

The work done by the net force equals the change in kinetic energy: (1/2)(1500 kg)(20 m/s)² = 300,000 J. Average power is work divided by time: 300,000 J / 10 s = 30,000 W.

Question 35Mechanics

A 2.0-kilogram block is pulled across a horizontal floor by a rope that makes a 30° angle above the horizontal with a tension of 10 N. If the block moves 3.0 meters horizontally, how much work is done by the tension?

  1. 15 J
  2. 30 J
  3. 0 J
  4. 26 J Correct answer.

Explanation

Work is the product of the force component in the direction of motion and the displacement. The horizontal component of the tension is (10 N) cos 30° ≈ 8.66 N, so the work is (8.66 N)(3.0 m) ≈ 26 J.

Question 36Mechanics

A 0.50-kilogram ball moving horizontally at 4.0 m/s strikes a wall and bounces back horizontally at 3.0 m/s. What is the magnitude of the impulse delivered to the ball by the wall?

  1. 1.5 kg·m/s
  2. 3.5 kg·m/s Correct answer.
  3. 0.50 kg·m/s
  4. 2.0 kg·m/s

Explanation

Impulse equals the change in momentum. Taking the initial direction as positive, the initial momentum is (0.50 kg)(4.0 m/s) = 2.0 kg·m/s and the final momentum is (0.50 kg)(−3.0 m/s) = −1.5 kg·m/s. The change is −1.5 − 2.0 = −3.5 kg·m/s, so the magnitude is 3.5 kg·m/s.

Question 37Mechanics

A 1200-kilogram car rounds a flat curve of radius 50 meters at a constant speed of 15 m/s. What is the magnitude of the centripetal acceleration of the car?

  1. 4.5 m/s² Correct answer.
  2. 3.0 m/s²
  3. 0.22 m/s²
  4. 0.30 m/s²

Explanation

Centripetal acceleration is given by a = v²/r. Substituting v = 15 m/s and r = 50 m gives (15)²/50 = 225/50 = 4.5 m/s².

Question 38Mechanics

A 5.0-kilogram block is pulled across a horizontal floor by a 20-newton force applied at 30° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.20. Using g = 9.8 m/s², what is the acceleration of the block?

  1. 1.0 m/s² Correct answer.
  2. 2.4 m/s²
  3. 0.50 m/s²
  4. 1.6 m/s²

Explanation

The horizontal component of the applied force is 20 cos 30° ≈ 17.3 N. The normal force is mg minus the vertical component of the applied force: (5.0)(9.8) − 20 sin 30° = 49 − 10 = 39 N. Friction is (0.20)(39) = 7.8 N, so the net horizontal force is about 17.3 − 7.8 = 9.5 N. Dividing by mass gives a ≈ 9.5/5.0 ≈ 1.9 m/s², closest to 1.6 m/s².

Question 39Mechanics

A 2.0-kilogram object moving at 6.0 m/s collides head-on with a 4.0-kilogram object moving at 2.0 m/s in the opposite direction. After the collision, the 2.0-kilogram object moves at 1.0 m/s in its original direction. What is the velocity of the 4.0-kilogram object?

  1. 0.50 m/s in its original direction
  2. 1.5 m/s opposite to its original direction
  3. 3.0 m/s opposite to its original direction
  4. 1.5 m/s in its original direction Correct answer.

Explanation

Total initial momentum is (2.0)(6.0) + (4.0)(−2.0) = 12 − 8 = 4.0 kg·m/s. After collision, the 2.0-kg object has momentum (2.0)(1.0) = 2.0 kg·m/s. The 4.0-kg object must have momentum 4.0 − 2.0 = 2.0 kg·m/s, so its velocity is 2.0/4.0 = 0.50 m/s in the original direction.

Question 40Mechanics

A 3.0-kilogram block slides from rest down a frictionless incline from a height of 5.0 meters. Using g = 9.8 m/s², what is the speed of the block at the bottom?

  1. 7.0 m/s
  2. 5.0 m/s
  3. 9.9 m/s Correct answer.
  4. 14 m/s

Explanation

Conservation of energy gives mgh = (1/2)mv², so v = √(2gh). Substituting h = 5.0 m and g = 9.8 m/s² gives v = √(98) ≈ 9.9 m/s.

Question 41Energy

A worker pushes a crate with a constant horizontal force of 40 N through a displacement of 5.0 m in the direction of the force. How much work does the worker do on the crate?

  1. 45 J
  2. 200 J Correct answer.
  3. 8.0 J
  4. 0 J

Explanation

Work equals the product of force and displacement when they point in the same direction: W = Fd = (40 N)(5.0 m) = 200 J.

Question 42Energy

A book is carried horizontally at constant speed across a room. What is the work done by the upward normal force on the book during this motion?

  1. Negative, because gravity acts downward
  2. Positive, because the book moves
  3. Equal to the weight times the distance moved
  4. Zero, because the force is perpendicular to the displacement Correct answer.

Explanation

Work depends on the component of force along the displacement: W = Fd cos θ. Here the normal force is vertical while the displacement is horizontal, so θ = 90° and cos 90° = 0, giving zero work.

Question 43Energy

Which quantity is defined as the rate at which work is done?

  1. Impulse
  2. Power Correct answer.
  3. Momentum
  4. Kinetic energy

Explanation

Power is the time rate of doing work, P = W/t, measured in watts (joules per second). The other quantities describe motion, force over time, or energy stored in motion.

Question 44Energy

A moving car has kinetic energy KE. If its speed is doubled while its mass stays the same, what is its new kinetic energy?

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

Explanation

Kinetic energy is KE = ½mv², so it is proportional to the square of speed. Doubling v multiplies KE by 2² = 4.

Question 45Energy

A 2.0 kg object is lifted 3.0 m at constant speed near Earth's surface. Taking g = 9.8 m/s², how much gravitational potential energy does it gain?

  1. 5.9 J
  2. 59 J Correct answer.
  3. 6.5 J
  4. 29 J

Explanation

Gravitational potential energy change is ΔPE = mgΔh = (2.0 kg)(9.8 m/s²)(3.0 m) ≈ 59 J.

Question 46Energy

Which statement best describes the law of conservation of energy?

  1. The total energy of a system always increases over time
  2. Energy is always destroyed as heat during any process
  3. Energy can be transformed between forms but the total amount stays constant in a closed system Correct answer.
  4. Energy can be created when work is done

Explanation

In an isolated system, energy may change form (for example, kinetic to potential or to thermal energy) but the total amount remains constant. Energy is neither created nor destroyed.

Question 47Energy

A spring with spring constant k = 200 N/m is compressed 0.10 m from its equilibrium length. How much elastic potential energy is stored in the spring?

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

Explanation

Elastic potential energy is PE = ½kx² = ½(200 N/m)(0.10 m)² = ½(200)(0.010) = 1.0 J.

Question 48Energy

A ball is dropped from rest from a height h and falls freely (ignore air resistance). Just before it hits the ground, its kinetic energy equals which quantity?

  1. mgh Correct answer.
  2. zero
  3. mg/h
  4. ½mgh

Explanation

All the gravitational potential energy mgh converts into kinetic energy as the ball falls, so KE = mgh just before impact.

Question 49Energy

A force of 50 N is applied to a box at an angle of 60° above the horizontal, and the box moves 4.0 m horizontally. How much work does the force do? (cos 60° = 0.50)

  1. 173 J
  2. 0 J
  3. 100 J Correct answer.
  4. 200 J

Explanation

Only the horizontal component of the force does work: W = Fd cos θ = (50 N)(4.0 m)(0.50) = 100 J.

Question 50Energy

A 1500 kg car accelerates from rest to 20 m/s. How much net work is done on the car?

  1. 150,000 J
  2. 30,000 J
  3. 300,000 J Correct answer.
  4. 600,000 J

Explanation

By the work-energy theorem, net work equals the change in kinetic energy: W = ½mv² = ½(1500 kg)(20 m/s)² = ½(1500)(400) = 300,000 J.