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MDCAT · Physics

Mechanics: Forces, Motion and Work MCQs for MDCAT

99 practice MCQs on kinematics, Newton's laws, momentum, work and energy, circular motion, rotation, gravitation, vectors and units. Every question has the correct answer and an explanation.

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Mechanics: Forces, Motion and Work is part of the MDCAT Physics syllabus. This page collects Mechanics: Forces, Motion and Work MCQs in the style of past MDCAT papers — below are 12 of them with answers and explanations, and the full set of 99 questions is in the HighYield MDCAT QBank.

This chapter covers: kinematics, Newton's laws, momentum, work and energy, circular motion, rotation, gravitation, vectors and units.

Mechanics: Forces, Motion and Work MCQs with answers

Q1.When the dimensions of both sides of an equation are equal, then the equation is said to be

  • A.Simultaneous
  • B.Homologous
  • C.Instantaneous
  • D.Quadratic
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Correct answer: B. Homologous

When both sides of an equation have identical dimensions, the equation is said to be homologous (dimensionally consistent), a key check used in dimensional analysis.

Q2.Using the relation for the magnifying power L o , M = 1 + d/f, if f = 5 cm and d = 25 cm then M will be:

  • A.5
  • B.7
  • C.6
  • D.8
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Correct answer: C. 6

Using the magnifying power formula M = 1 + D/f with D = 25 cm and f = 5 cm, M = 1 + 25/5 = 6.

Q3.0.1 kg is equivalent to the energy of:

  • A.9x1016
  • B.9x1015
  • C.8×106
  • D.3x107
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Correct answer: B. 9x1015

Option B is correct because 0.1 kg is equivalent to 9x1015 joules of energy, based on the mass-energy equivalence principle stated by Einstein's famous equation E=mc2.

Q4.Ignoring details associated with friction, extra forces exerted by arm and leg muscles, and other factors, we can consider a pole vault as the conversion of an athlete's running kinetic energy to gravitational potential energy. If an athlete is to lift his body 5m during a vault, what speed must he have when he plants his pole?

  • A.5 m/s
  • B.10 m/s
  • C.15 m/s
  • D.20 m/s
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Correct answer: B. 10 m/s

Given: Distance = S = 5m Acceleration = a = g = 10ms-2 Initial velocity = Vi = 0 ms-1 Final velocity = Vf = ? Using third equation of motion: 2aS = Vf² - Vi² Vf² = 2aS + Vi² = 2 × 10 × 5 + (0)² Vf² = 100 Taking square root on both sides, we get: Vf = 10ms-1

Q5.The velocity-time plot for a moving particle shows a straight line. This means:

  • A.The particle has a constant acceleration
  • B.The particle has never turned around
  • C.The particle has zero displacement
  • D.The data in insufficient
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Correct answer: A. The particle has a constant acceleration

The correct statement is A. The particle has a constant acceleration. A straight line on a velocity-time plot typically indicates that the particle is undergoing constant acceleration.

Q6.If a particle is moving with uniform circular motion, then:

  • A.Velocity and acceleration are antiparallel
  • B.Velocity and acceleration are parallel
  • C.Velocity and acceleration are perpendiculars
  • D.Zero acceleration
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Correct answer: C. Velocity and acceleration are perpendiculars

The correct option is C, "Velocity and acceleration are perpendicular." This option is correct. In uniform circular motion, the velocity and acceleration vectors are always perpendicular to each other. The velocity is tangential to the circular path, while the acceleration (centripetal acceleration) points toward the center of the circle.

Q7.The centripetal acceleration of an object moving along a circle of radius 'r' with an angular speed 'w' is given by the formula:

  • A.a= r W2
  • B.a= r W
  • C.a = r² W
  • D.a= r² W²
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Correct answer: A. a= r W2

Centripetal acceleration increases with radius and is also affected more significantly when angular speed increases, as it is squared.

Q8.A body is moved through a displacement of 15m towards the North, the total displacement will be zero when the body covered the same displacement towards:

  • A.North
  • B.South
  • C.West
  • D.East
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Correct answer: B. South

Displacement refers to the change in position of an object and is a vector quantity, meaning it has both magnitude and direction. In this question, the body initially moves 15m towards the North, resulting in a positive displacement of 15m in that direction. To achieve a total displacement of zero, the body must move back the same distance in the opposite direction. Therefore, moving 15m towards the South will cancel out the initial Northward displacement, resulting in a total displacement of zero.Options North, West, and East do not negate the initial displacement of 15m North, hence they do not result in zero total displacement.

Q9.The angular speed in radian/hour for the daily rotation of our earth is?

  • A.π/12 radian/hour
  • B.12π radian/hour
  • C.24π radian/hour
  • D.π/24 radian/hour
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Correct answer: A. π/12 radian/hour

The correct answer is π/12 radian/hour. To find the angular speed in radians, we recognize that one full rotation (2π radians) occurs in 24 hours. Thus, the angular speed can be calculated as:Angular speed = Total radians per rotation / Time in hours = 2π / 24 = π/12 radian/hour.

Q10.A boy raises an object of mass 10Kg vertically above his head through a height of 5m. The potential energy stored in the body is:_

  • A.98 J
  • B.49J
  • C.490 J
  • D.980 J
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Correct answer: C. 490 J

The potential energy (PE) stored in an object is calculated as:PE = m × g × hwhere:m = mass of the object (10 kg)g = acceleration due to gravity (approximately 9.8 m/s²)h = height through which the object is raised (5 m)Substituting the values, we get:PE = 10 kg × 9.8 m/s² × 5 m= 490 J

Q11.Newton's 3rd law of motion is correlated with

  • A.law of conservation of energy
  • B.momentum
  • C.law of conservation of velocity
  • D.All
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Correct answer: B. momentum

Newton's 3rd law of motion states: "For every action, there is an equal and opposite reaction." This means that when object A exerts a force on object B, object B exerts an equal and opposite force on object A.This law is closely related to the law of conservation of momentum, which states that the total momentum of a closed system remains constant over time. When object A exerts a force on object B, it transfers momentum to object B, and vice versa. The equal and opposite forces result in equal and opposite changes in momentum, conserving the total momentum of the system.

Q12.In projectile motion, the horizontal component of acceleration is:

  • A.Zero
  • B.9.8m/s²
  • C.19.6m/s
  • D.4.9m/s²
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Correct answer: A. Zero

In projectile motion, the horizontal component of acceleration is zero. This is because no external force acts horizontally (assuming air resistance is negligible), so the horizontal velocity remains constant throughout the motion. The only acceleration is the vertical component, which is due to gravity.

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