15 GAMSAT Physics sample questions for free!

Gold Standard GAMSAT has over 6000 practice problems with worked solutions, so we thought we should start sharing some! Don't worry, we will start off with the basics and then work up to GAMSAT-level reasoning.

"Practice isn't the thing you do once you're good. It's the thing you do that makes you good."

GAMSAT Maths and Physics Section 3


The GAMSAT physics sample questions below come in 2 varieties: 1) Foundational questions (NOT GAMSAT-level) which are intended to ensure that you have a basic understanding of physics, which is helpful for more reasoning questions; 2) GAMSAT-level physics questions: Questions 4, 8, 11, 13 and 15. For long-form practice questions, consider our free full-length GAMSAT practice test. Worked solutions for the questions on this webpage can be found below Question 15.

Ideally, you would time yourself to complete these 15 questions. You have 30 minutes. Answers and worked solutions will follow. Oh, and of course, no calculators.

Gold Standard

If a metric ton is 1000 kg, to pull the door open at 40,000 feet is equivalent to moving how many metric tons?

Answer Key:
1. C
2. D
3. D
4. C
5. B
6. D
7. D
8. B
9. D
10. C
11. B
12. C
13. D
14. A
15. B

*Gold Standard GAMSAT textbook Reference

1) If you are adding these 2 vectors, let's consider the 2 extreme situations (i.e. the highest possible value and the lowest possible value): (1) they are pointing in exactly the same direction so the sum is 11 + 5 = 16 N; (2) they are pointing in exactly the opposite directions so the sum is 11 + (-5) = 6 N. Anything from 6 N to 16 N is possible but nothing above or below.
*PHY 1.1

2) Newton's 1st law states that an object in motion will stay in motion at constant velocity unless acted upon by an unbalanced force. For something to change direction or accelerate there must be an unbalanced force acting on it.
*PHY 4.2

3) If the direction changes, then all vectors change: momentum, displacement and velocity. Kinetic energy is a scalar and does not depend on direction. As long as the speed remains the same, the kinetic energy remains the same.
*PHY 1.1, 5.1, 5.2

4) Imagine that you needed to run down a hallway but you had 2 choices: (1) a very narrow hallway with a certain density of students there, just milling around OR (2) a wide hallway with the SAME density of students. Think for a moment. If you run down a wide hallway with the same density, you always have more options to choose from: left, right, sideways and so you can get through faster. It's the same with the electrons going through these 2 wires, same density but one with a greater radius.

Actually, a wire is like a tiny cylinder and if you cut a wire and look at its edge, the shape is a tiny circle. So it's not just a matter of the radius that is bigger, it's actually the area of a circle that is bigger. So we can summarize by saying: as the area increases, resistance decreases. Or, in "math-speak", resistance is inversely proportional to the cross-sectional area.
Now we have our first basic high school equation that must be memorized!
The resistance (R) is inversely proportional to the cross-sectional area (A), where
Acircle = π(radius)2
If radius1 = 2 x radius2, then A1 = (2A2)2 = 4A2 and R1 = 1/4 R2, giving us a ratio of R1:R2 --> 4:1.
*PHY 10.2 (B: PHY 3.3)

5) You would be expected to know: Newton's Law of Gravitation F = Gm1m2/r2 (you would never perform a complete calculation on the GAMSAT using values for each of the unknowns in the equation because there is just not enough time! However, you must know the relationship so that you can make assessments as to the relative changes of the variables in the equation).

When the object is moved 1 Earth radius away from the Earth then r is now 2r (on the surface of the earth the object is 1 Earth radius away from the center of the Earth, when the object is moved 1 radius away the distance is now r + r = 2r).

When you plug (2r)2 into the denominator of the equation it becomes 4r2. Thus when you double the distance, you quarter the force (same relationship with Coulomb's Law and point charges). So if we start with a force of 60 N, then we end with 15 N. Not much to calculate!
*PHY 2.4

6) This is a common exam question testing your understanding of Newton's 3rd Law: for every action, there is an equal and opposite reaction. When you push a couch, the couch is pushing back against your body with an equal force. The mass in the question exerts a 10 N force back on the spring because action/reaction forces are equal.
*PHY 2.3

7) The tension in the ropes is largest when the ropes are vertical because at that moment, the centripetal force is maximum and in the same direction as gravity. Clearly, the speed of the child is maximum at the point of lowest (gravitational) potential energy; therefore, the centripetal force is also maximum at that point.

Centripetal force = mv2/r = (20 x 3 x 3)/3 = 60 N
Weight = 200 N
Maximum force = 200 + 60 = 260 N
Maximum tension in each rope = 260/2 = 130 N
*PHY 3.3

8) First let's calculate the specific gravity (SG) of wood which is simply the density of wood divided by the density of water which is 400/1000 = 0.4.
SG is equal to the fraction of the height of a buoyant (= floating) object below the surface. So we know the fraction is 0.4 and now we need to apply that to the total "height" which is the depth of the wood 10 cm. So 0.4 x 10 = 4 cm.
*PHY 6.1.1, 6.1.2

9) The right-hand rule describes the direction of the magnetic field: as you "grab" the wire with your thumb in the direction of the flow of current, the tips of your fingers trace the direction of the magnetic field which encircles the wire. See PHY 9.2.3 of the Gold Standard GAMSAT book.
*PHY 9.2.2, 9.2.3

10) v = vo + gt
t = (v-vo)/g = 10/10 = 1 s
total time in air (to go up and back down) = 2t = 2 s
And, s = vot + (g/2)(t2)
= (10 x 1) - (10/2)(1)
= 5 m (max. height)

Yes, there are some very basic physics equations that you should know for the GAMSAT and we have listed them all here: www.gamsat-prep.com/section3-physics-formula-list/
*PHY 1.6, 2.6

11) While in the area between the plates, the negatively charged electron is attracted to the positive plate and so it would bend "upwards". But after leaving the plates, there is no more net force acting on the electron. Therefore, the electron would continue in motion in a straight line according to Newton's first law (inertia).

Oh, btw, the real GAMSAT: only 4 answer choices for all multiple choice questions. We just had to make this question 20% better but we promise not to do it again!! : )
*PHY 9.1.3, 4.2

12) You can draw a free-body diagram of the person:

If upwards is positive, and we know that the sum of forces is equal to ma from Newton's 2nd Law, we get:
Fscale - weight = ma
The scale reading is less than his normal weight, so the acceleration is negative; i.e., down. Downward acceleration means either something speeds up while moving downward, or slows down while moving upward.
*PHY 2.2, 3.4

13) Key points: (1) pressure is defined as force per unit area (just apply some pressure to the table in front of you and you'll would have to say that you are applying a certain force over a particular area); (2) dimensional analysis: kilo means 1000 in the SI system and pascals or Pa, being force/area, has the equivalent units of N/m2.

The difference in pressure between inside and outside the cabin at 40,000 feet is: 75 – 20 = 55 kPa = 55 000 Pa or N/m2 of pressure pushing on the door.
Pressure: P = F/A
The area of the door is 2 x 1 = 2m2
So F = PA = (55 000)(2) = 110 000 N
Thus weight mg = 110 000 meaning m = 11 000 kg or 11 metric tons.
*PHY 6.1.2

14) Because no forces act perpendicular to the incline except for the normal force and the perpendicular component of weight, and there is no acceleration perpendicular to the incline, the normal force is equal to the perpendicular component of weight, which is mg cosθ. As the angle increases, the cosine of the angle decreases. This decrease is nonlinear because a graph of the normal force N vs. θ would show a curve, not a line.
*PHY 3.2, 3.2.1

15) The 10 A of current splits into two paths. The path through R2 takes 7 A, so the path through R3 takes 3 A (this is Kirchoff's First Law, the junctional theorem). Now use Ohm's law, where the voltage V is equal to the current I times the resistance R (Ohm's law is a frequent visitor in the real GAMSAT).
V = IR
V = (3 A) (10 Ω) = 30 V.
*PHY 10.1, 10.2, 10.3

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