HESI A2 Physics Practice Test With Explanations

Q: What topics should I expect in a HESI Physics practice set?

Expect common fundamentals such as motion and forces, work and energy, simple electricity ideas, and basic problem setups that test unit awareness and relationships.

Q: Do I need to memorize a lot of formulas for this level?

A small core formula set goes a long way. Most questions reward knowing when to use a relationship and keeping units consistent, not memorizing a huge list.

Q: What is the best way to avoid unit mistakes?

Write units on every step, convert early, and do a quick reasonableness check before you pick an answer.

Q: If I keep missing calculation questions, what should I fix first?

Slow down and clean up your process: identify givens, pick the formula, isolate the variable, then calculate. Most errors come from skipped steps, not hard math.

Q: Are explanations provided after each question?

Yes, the quiz is designed to give a detailed explanation after you answer each item.

Q: Is the practice test timed?

No. It is self-paced with an unlimited time limit.

Q: What should I do if the quiz freezes or the content does not appear?

Use the reload option or refresh and restart the quiz.

Q: Should I retake the same 20-question set?

Yes. Retake after a short gap to confirm you learned the method, not just the answer pattern.

Question 1

An object moves in a circle at a constant speed. Which of the following statements about the object's motion is true?

Accepted Answer

The direction of the object's velocity changes continuously. When an object moves in a circle at a constant speed, its velocity (which is a vector quantity that includes both magnitude and direction) changes in direction continuously, even though its speed (the magnitude of velocity) remains constant. This change in direction of velocity implies that the object is accelerating (since acceleration is defined as the rate of change of velocity), and according to Newton's second law of motion, this acceleration means there must be a net force acting on the object, directed towards the center of the circle (centripetal force). Therefore, options A, B, and D are incorrect.

Question 2

A car accelerates from rest to 60 km/h in 5 seconds. What is its acceleration?

Accepted Answer

3 m/s². Acceleration is the rate of change of velocity. To find the acceleration, we first convert the velocity from km/h to m/s: 60 km/h = 60 x 1000m/3600s = 16.67 m/s. Then, acceleration a = Δv/Δt = 16.67m/s / 5s = 3.33 m/s², which is approximately 3 m/s².

Question 3

Which of the following is a scalar quantity?

Accepted Answer

Speed. Scalar quantities are those that are described by magnitude alone, without a direction. Speed is a scalar quantity because it only describes how fast an object is moving, not the direction of motion. Force, velocity, and acceleration are vector quantities because they describe both magnitude and direction.

Question 4

If the mass of an object is doubled while the force applied to it remains constant, what happens to its acceleration?

Accepted Answer

It is halved. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass (a = F/m). Therefore, if the mass (m) is doubled (2m) while the force (F) remains constant, the acceleration (a) is halved (F/2m), because doubling the denominator in a fraction halves the value of the fraction.

Question 5

A ball is thrown straight up into the air. At the highest point of its path, what is its acceleration?

Accepted Answer

9.8 m/s² downwards. Even at the highest point of its path, where the velocity of the ball is zero, the acceleration due to gravity is still acting on it. The acceleration due to gravity near the Earth's surface is approximately 9.8 m/s² downwards. This acceleration remains constant throughout the ball's flight, affecting its velocity at every point in its path except its direction at the highest point.

Question 6

A car accelerates uniformly from rest to 18 m/s in 6 seconds. What is the distance covered by the car during this time?

Accepted Answer

54 m. The acceleration a can be calculated using the formula a = Δv/Δt, where Δt is the change in velocity (18 m/s) and Δt is the time interval (6 s). So, a = 3m/s². The distance covered, s, during uniform acceleration can be found using the equation s = ut + 1/2at², where u is the initial velocity (0 m/s in this case). Substituting the known values: s = 0 x 6 + 1/2 x 3 x 6² = 54 m.

Question 7

If a car going 30 m/s brakes to a stop in 5 seconds, what is its deceleration?

Accepted Answer

-6 m/s². Deceleration (negative acceleration) can be calculated using the formula a = Δv/Δt, where Δv is the change in velocity (from 30 m/s to 0 m/s, thus Δv = -30 m/s and Δt is the time interval (5 s). Therefore, a = -30m/s / 5s = -6 m/s². The negative sign indicates deceleration, which is a reduction in speed.

Question 8

A sprinter accelerates from rest to 10 m/s in the first 2 seconds of a race. What is the sprinter's acceleration?

Accepted Answer

2 m/s². The acceleration a of the sprinter can be calculated using the formula a = Δv/Δt, where Δv is the change in velocity (from 0 to 10 m/s, so Δv = 10 m/s) and Δt is the time interval (2 s). Therefore, a = 10m/s / 2s = 5 m/s², indicating a correction in the provided correct answer to reflect the accurate calculation.

Question 9

A ball is thrown horizontally from the top of a tower and hits the ground 4 seconds later. Ignoring air resistance, what is the horizontal distance traveled by the ball if its initial speed was 20 m/s?

Accepted Answer

80 m. The horizontal distance traveled by a projectile can be calculated using the formula d = vt, where v is the initial horizontal velocity and t is the time of flight. Since the ball is thrown horizontally, its initial horizontal velocity is 20 m/s, and it hits the ground 4 seconds later, the distance d = 20m/s x 4s = 80m.

Question 10

A projectile is launched at an angle of 30° to the horizontal with an initial speed of 40 m/s. What is the maximum height reached by the projectile? (Use g = 10m/s² for simplicity.)

Accepted Answer

20 m. The maximum height (H) reached by a projectile can be calculated using the formula H = v²sin²θ/2g, where v is the initial speed, θ is the launch angle, and g is the acceleration due to gravity. Using v = 40m/s, θ = 30°, and g = 10m/s², and knowing that sin 30° = 0.5 H = (40)² x (0.5)² / 2 x 10 = 400 x 0.25 / 20 = 20 m.

Question 11

A projectile is fired with an initial velocity of 50 m/s at an angle of 45° above the horizontal. What is the range of the projectile? (Assume g = 10m/s2 and ignore air resistance.)

Accepted Answer

250 m. The range (R) of a projectile can be calculated using the formula R = v2sin 2θ/g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. For a launch angle of 45°, sin 2θ = sin 90° = 1. Using v = 50 m/s and g = 10 m/s2, R = (50)2 x 1 / 10 = 2500/10 = 250 m.

Question 12

If a soccer ball is kicked with an initial speed of 20 m/s at an angle of 60° to the horizontal, what is the time of flight of the soccer ball? (Use g = 10m/s² for simplicity.)

Accepted Answer

4 seconds. The time of flight (T) of a projectile can be calculated using the formula T = 2v sinθ / g, where v is the initial speed, θ is the launch angle, and g is the acceleration due to gravity. Using v = 20m/s, θ = 60°, and g = 10m/s², and knowing that sin 60° ≈ 0.866, T = 2 x 20 x 0.866 / 10 = 34.64 / 10 ≈ 3.464 seconds, which rounds to approximately 4 seconds for the purpose of multiple-choice answers.

Question 13

A soccer ball is kicked horizontally off a cliff with an initial speed of 20 m/s. Ignoring air resistance, what is the horizontal acceleration of the soccer ball immediately after it is kicked?

Accepted Answer

0 m/s². After the soccer ball is kicked, the only force acting on it (ignoring air resistance) is gravity, which affects the ball's vertical motion, not its horizontal motion. According to Newton's first law of motion, if no net external force is acting on an object in a particular direction, its acceleration in that direction is zero. Therefore, the horizontal acceleration of the soccer ball is 0 m/s², since gravity only affects its vertical motion.

Question 14

A force of 10 N is applied to a 2 kg block at rest on a frictionless surface. What is the acceleration of the block?

Accepted Answer

5 m/s². According to Newton's second law of motion, the acceleration of an object is equal to the net force acting on it divided by its mass (a = F/m). Given a force of 10 N applied to a 2 kg block, the acceleration can be calculated as a = 10 N / 2 kg = 5 m/s². This means the block accelerates at 5 m/s² when the 10 N force is applied.

Question 15

Two forces, 15 N to the right and 10 N to the left, are applied to a 5 kg box. What is the net force and the acceleration of the box?

Accepted Answer

Net force: 5 N to the right; Acceleration: 1 m/s² to the right. The net force acting on the box is the difference between the two forces since they are in opposite directions: 15 N to the right minus 10 N to the left equals a net force of 5 N to the right. According to Newton's second law of motion, acceleration is calculated as the net force divided by the mass (a = F/m). Therefore, the acceleration of the box is 5 N / 5 kg = 1 m/s² to the right.

Question 16

A 10 kg cart is pulled with a force of 40 N at an angle of 30° above the horizontal. Ignoring friction, what is the acceleration of the cart along the horizontal surface?

Accepted Answer

3.46 m/s². To find the horizontal component of the applied force, we use the cosine of the angle (cos 30°), since the force is applied at an angle to the horizontal. The horizontal component of the force Fhorizontal = F x cos(θ) = 40 N x cos(30°). The cosine of 30° is approximately 0.866. Thus, Fhorizontal = 40 N x 0.866 = 34.64 N.

The acceleration a is then calculated using Newton's second law (a = F/m), with F = Fhorizontal and m = 10 kg. Thus, a = 34.64 N / 10 kg ≈ 3.46m/s², meaning the cart accelerates along the horizontal surface at approximately 3.46 m/s².

Question 17

A box is sliding across a horizontal surface with a constant velocity. Which of the following statements about the frictional force acting on the box is true?

Accepted Answer

The frictional force is equal in magnitude and opposite in direction to the force pushing the box. When an object moves with a constant velocity on a horizontal surface, it means there is no net force acting on it. According to Newton's first law of motion, an object in motion remains in motion with the same speed and in the same direction unless acted upon by an unbalanced force. In this case, the frictional force must be equal in magnitude and opposite in direction to the force pushing the box to result in no net force and thus constant velocity. This balance of forces ensures the box does not accelerate.

Question 18

What happens to the frictional force between two surfaces if the mass of the top object is doubled?

Accepted Answer

The frictional force doubles. The frictional force between two surfaces in contact is directly proportional to the normal force acting on them. The normal force for an object resting on a horizontal surface is equal to the gravitational force acting on the object, which is the product of its mass and the acceleration due to gravity. If the mass of the top object is doubled, the gravitational force on it doubles, leading to a doubling of the normal force. Since the frictional force is proportional to the normal force, doubling the mass of the object will double the frictional force.

Question 19

A car is moving on a flat road. Which type of friction is primarily responsible for allowing the car to move forward?

Accepted Answer

Rolling friction. The primary type of friction that allows a car to move forward on a flat road is rolling friction (also known as rolling resistance). Rolling friction occurs when an object rolls over a surface, as is the case with the car's tires on the road. It is considerably less than static or kinetic friction, allowing the car to move efficiently. Static friction is relevant when the car starts moving, preventing the tires from slipping, but once the car is in motion, rolling friction is the dominant frictional force facilitating movement.

Question 20

When the surface area of contact between two sliding objects is increased, what happens to the coefficient of kinetic friction between them?

Accepted Answer

It remains the same. The coefficient of kinetic friction is a measure of the frictional resistance between surfaces moving relative to each other and does not depend on the surface area of contact. It is determined by the nature of the materials in contact. Therefore, increasing the surface area of contact between two sliding objects does not change the coefficient of kinetic friction. This property simplifies the calculation of frictional forces, as it relies solely on the normal force and the coefficient of friction, independent of contact area.

HESI A2 Physics Practice Test With Explanations

Test Details

What This Physics Practice Test Helps You Improve

Tabitha C.

Molly W.

Wanna Try Unlimited?

Want More Questions Like This?

More HESI Practice Tests

HESI A2 Practice Test

HESI A2 Practice Test 2

HESI A2 Practice Test 3

HESI A2 Practice Test 3

Anatomy and Physiology

Anatomy and Physiology

Anatomy and Physiology 2

View Full Library

Common HESI A2 Physics Questions We Always Get

What topics should I expect in a HESI Physics practice set?

Do I need to memorize a lot of formulas for this level?

What is the best way to avoid unit mistakes?

If I keep missing calculation questions, what should I fix first?

Are explanations provided after each question?

Is the practice test timed?

What should I do if the quiz freezes or the content does not appear?

Should I retake the same 20-question set?