All Free Physics MCQs with Answers

11. A body of mass 2 kg moving at 3 m s-1 collides and sticks to a stationary body of mass 4 kg. The common velocity after impact is

• A. 0.5 m s-1
• B. 1 m s-1
• C. 1.5 m s-1
• D. 2 m s-1

Explanation: Momentum is conserved, so the initial momentum of 2 times 3, which is 6 kg m s-1, is shared by the combined mass of 6 kg, giving a velocity of 1 m s-1. This is a perfectly inelastic collision, so kinetic energy is not conserved even though momentum is. Only the moving body contributes momentum, since the other starts at rest.

12. In an elastic collision, which quantity is conserved in addition to momentum?

• A. Kinetic energy
• B. Mass alone
• C. Velocity of each body
• D. Potential energy

Explanation: An elastic collision is defined by the conservation of kinetic energy as well as momentum, and collisions between gas molecules are treated this way. In an inelastic collision momentum is still conserved but some kinetic energy is converted into heat, sound or deformation. Individual velocities change in every collision.

13. A car accelerates uniformly from rest to 20 m s-1 in 8 s. The distance covered is

• A. 40 m
• B. 80 m
• C. 160 m
• D. 320 m

Explanation: With uniform acceleration from rest the average velocity is half the final velocity, which is 10 m s-1, and over 8 s that gives 80 m. The same answer comes from finding a as 20 divided by 8, which is 2.5 m s-2, then using S equal to half a t squared. Using the final velocity of 20 m s-1 for the whole 8 s gives the wrong answer of 160 m.

14. The area under a velocity-time graph represents

• A. acceleration
• B. displacement
• C. force
• D. momentum

Explanation: Multiplying velocity by time gives displacement, so the area enclosed between the curve and the time axis is the displacement travelled. The gradient of the same graph gives acceleration. On a displacement-time graph, by contrast, the gradient is velocity and the area has no useful physical meaning.

15. Work done by a force is zero when the angle between the force and the displacement is

• A. 0 degrees
• B. 45 degrees
• C. 90 degrees
• D. 180 degrees

Explanation: Work is the scalar product Fd cos theta, and cos 90 degrees is zero, so a force perpendicular to the displacement does no work. This is why the centripetal force in uniform circular motion does no work and the speed stays constant. At 180 degrees the work is negative rather than zero, as with friction.

16. A porter carries a suitcase horizontally across a platform at constant speed. The work done by the porter against gravity is

• A. equal to mgh
• B. zero
• C. equal to the weight times the distance walked
• D. negative

Explanation: The upward force supporting the suitcase is perpendicular to the horizontal displacement, so the work done against gravity is zero. The porter still expends metabolic energy holding the case up, but that is not mechanical work on the suitcase, which is the standard point of this question. Work against gravity appears only when the height changes.

17. The kinetic energy of a body is doubled when its

• A. mass is halved
• B. speed is doubled
• C. speed is increased by a factor of the square root of two
• D. mass is doubled and speed halved

Explanation: Kinetic energy is half m v squared, so it scales with the square of the speed. Multiplying the speed by root 2 multiplies the energy by 2, whereas doubling the speed multiplies the energy by 4. The last option leaves energy at half its original value, since doubling m and quartering v squared gives a net factor of one half.

18. A body of mass 5 kg moves at 4 m s-1. Its kinetic energy is

• A. 10 J
• B. 20 J
• C. 40 J
• D. 80 J

Explanation: Kinetic energy is half times 5 times 4 squared, which is half times 5 times 16, giving 40 J. The common slip is to forget to square the velocity, which gives 10 J, or to forget the factor of one half, which gives 80 J.

19. The work-energy theorem states that the net work done on a body equals

• A. its total mechanical energy
• B. the change in its kinetic energy
• C. the change in its potential energy
• D. its momentum

Explanation: The work-energy theorem equates net work with the change in kinetic energy, which is why a braking force doing negative work brings a car to rest. It holds whether the force is constant or varying. Changes in potential energy relate to the work done by conservative forces specifically, which is a narrower statement.

20. Power is correctly defined as

• A. the total work done
• B. the rate of doing work
• C. work done times time taken
• D. force times displacement

Explanation: Power is work done per unit time, measured in watts, where one watt is one joule per second. For a body moving at constant velocity it can also be written as force times velocity. Force times displacement is work itself, not the rate at which it is done.