Kinematics and Kinetics MCQ Quiz - Objective Question with Answer for Kinematics and Kinetics - Download Free PDF

Last updated on Jun 17, 2025

Latest Kinematics and Kinetics MCQ Objective Questions

Kinematics and Kinetics Question 1:

The rate of decay of oscillations is known as

  1. Transmissibility
  2. Damping coefficient
  3. Logarithmic decrement
  4. Critical damping

Answer (Detailed Solution Below)

Option 3 : Logarithmic decrement

Kinematics and Kinetics Question 1 Detailed Solution

Explanation:

The Rate of Decay of Oscillations

  • The rate of decay of oscillations refers to the decrease in the amplitude of oscillatory motion over time due to energy dissipation. In mechanical systems, this decay is primarily caused by damping forces that convert the mechanical energy of oscillation into other forms, such as heat. The term used to quantify this rate of decay is known as the "Logarithmic Decrement."

Logarithmic Decrement:

  • Logarithmic decrement is a measure that quantifies the rate at which the amplitude of an oscillatory system decreases between successive cycles. It is commonly represented by the Greek letter δ (delta). This parameter is particularly useful in analyzing underdamped systems, where oscillations persist but gradually reduce in amplitude. Logarithmic decrement is mathematically expressed as:

Formula:

Logarithmic Decrement (δ) = ln(A₁/A₂)

Where:

  • A₁ = Amplitude of the first oscillation
  • A₂ = Amplitude of the second oscillation
  • ln = Natural logarithm

Working Principle: Logarithmic decrement is derived from the damping characteristics of the system. When a system is set into oscillatory motion, damping forces—such as viscous damping—cause the amplitude of the oscillation to decrease with each successive cycle. The logarithmic decrement provides a quantitative measure of this decay and helps engineers and analysts determine the damping efficiency of the system.

Applications:

  • Used in vibration analysis to evaluate the damping characteristics of structures and machinery.
  • Helpful in designing mechanical systems to ensure stability and mitigate excessive vibrations.
  • Applied in various engineering fields such as civil engineering, automotive engineering, and aerospace engineering to optimize system performance.

Kinematics and Kinetics Question 2:

The acceleration of a particle is expressed as a = 10 - x. The particle starts (time = 0) with no initial velocity at x = 0. What will be the velocity of the particle when acceleration is zero?

  1. 10 m/s
  2. 5 m/s
  3. 8 m/s
  4. 15 m/s

Answer (Detailed Solution Below)

Option 1 : 10 m/s

Kinematics and Kinetics Question 2 Detailed Solution

Concept:

We analyze the motion of a particle with variable acceleration to determine its velocity when the acceleration becomes zero.

Given:

  • Acceleration function:
  • Initial conditions: at

Step 1: Find the position when acceleration is zero

Set

Step 2: Relate acceleration to velocity and position

Using the chain rule, acceleration can be expressed as:

Separate variables and integrate:

Step 3: Integrate both sides

 

Kinematics and Kinetics Question 3:

A car starts from rest on a curved road of radius 200 m and attains a speed of 18 km/hr at the end of 60 seconds while travelling with constant tangential acceleration. What will be the normal acceleration of car after 30 seconds from the start?

  1. 0.0886 m/s2
  2. 0.03125 m/s2
  3. 0.083 m/s2
  4. 0.0252 m/s2

Answer (Detailed Solution Below)

Option 2 : 0.03125 m/s2

Kinematics and Kinetics Question 3 Detailed Solution

Concept:

When a body moves along a curved path with constant tangential acceleration, the total acceleration has two components:

1. Tangential acceleration due to change in speed.

2. Normal (centripetal) acceleration due to change in direction, given by:

Calculation: 

Given:

Final speed after 60 seconds: v = 18 km/h = 5 m/s

Radius of curve: r = 200 m

Time = 60 s, Initial speed = 0 (starts from rest)

Step 1: Calculate tangential acceleration

Step 2: Speed at 30 seconds:

Step 3: Normal acceleration

 

Kinematics and Kinetics Question 4:

Two guns (A and B) are pointed at each other, A upwards at an angle of 30° with horizontal and B at the same angle of depression as shown in figure. The guns are 40 m apart. If the gun A fires (shot) at the velocity of 350 m/s and gun B fires (shot) at the velocity of 300 m/s respectively at the same time. The shots meet at M. What will be the time of meeting after firing?

  1. seconds
  2.  seconds
  3.  seconds
  4.  seconds

Answer (Detailed Solution Below)

Option 4 :  seconds

Kinematics and Kinetics Question 4 Detailed Solution

Concept:

To find the time of meeting of the two shots, we consider the relative motion of the projectiles in the horizontal direction.

Given:

Distance between guns, d = 40 m

Velocity of shot from Gun A, vA = 350 m/s

Velocity of shot from Gun B, vB = 300 m/s

Angle of projection, 

Calculation:

Resolving velocities into horizontal components:

m/s

m/s

The relative velocity in the horizontal direction is:

m/s

Using the formula,

Approximating  ,

 seconds

Kinematics and Kinetics Question 5:

A sphere of mass 3 kg strikes at velocity of 5 m/s to another sphere of mass 2 kg, which is at rest. What would be their common velocity, if they move together after collision?

  1. 6 m/s
  2. 3 m/s
  3. 9 m/s
  4. 12 m/s

Answer (Detailed Solution Below)

Option 2 : 3 m/s

Kinematics and Kinetics Question 5 Detailed Solution

Concept:

When two bodies collide and move together after the collision, the principle of conservation of momentum is applied.

The equation for conservation of momentum is given by:

Where:

  • kg (mass of first sphere)
  • m/s (velocity of first sphere before collision)
  • kg (mass of second sphere)
  • m/s (velocity of second sphere before collision, since it is at rest)
  • is the common velocity after collision

Calculation:

Applying the conservation of momentum equation:

m/s

Top Kinematics and Kinetics MCQ Objective Questions

An object starts from rest at x = 0 m and t = 0 s. It moves with a constant acceleration of 2m/s2 along the x–axis. What is its average velocity between time 1 s and 5 s?

  1. 2 m/s
  2. 8 m/s
  3. 6 m/s
  4. 4 m/s

Answer (Detailed Solution Below)

Option 3 : 6 m/s

Kinematics and Kinetics Question 6 Detailed Solution

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Concept:

  • Average velocity = total displacement/ total time duration
  • Equation of motion:
  • v = u + at
  • v2 = u2 + 2as
  • s = ut + 1/2 at2

Calculation:

Given:

Time interval = 5 s & 1 s, Initial velocity u = 0, and, Acceleration a = 2 m/s2

 

​When an object starts from rest, then the total distance covered in time 1 sec and 5 sec is,

s = ut + 1/2 at2

Object is at rest, so, u = 0 m/s.

s2 - s1 = 24 m

Total time taken, t = t2 - t1 = 5 - 1 = 4 sec

Average velocity = total displacement/ total time duration

Average velocity = 24/4 = 6 m/s

Average velocity between time 1 s and 5 s = 6 m/s

A vehicle having rectilinear motion is moving with a velocity of 36 km/h and accelerates uniformly to 54 km/h over a distance of 125 m. What will be the time taken to cover this distance? 

  1. 5 seconds  
  2. 15 seconds 
  3. 20 seconds 
  4. 10 seconds  

Answer (Detailed Solution Below)

Option 4 : 10 seconds  

Kinematics and Kinetics Question 7 Detailed Solution

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Concept:

Rate of change of velocity is known as acceleration. Its unit is m/s2. It is a vector quantity.

a = change in velocity/time

Equations of motion:

  • v = u + at
  • v2 – u2 = 2as

Calculation:

Given:

u = 36 km/h = 10 m/s; S = 125 m ; v = 54 km/h = 15 m/s,  t = ?

v2 – u2 = 2as

v = u + at

A body moves with a speed of 10 m/s in the curved path of 25 m radius of curvature. If the tangential acceleration is 3 m/s2, then total acceleration for the body will be:

  1. 3.3 m/s2
  2. 4 m/s2
  3. 5 m/s2
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 5 m/s2

Kinematics and Kinetics Question 8 Detailed Solution

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CONCEPT:

Centripetal Acceleration (ac): 

  • Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
  • It always acts on the object along the radius towards the center of the circular path.
  • The magnitude of centripetal acceleration,

Where v = velocity of the object and r = radius

Tangential acceleration (at):

  • It acts along the tangent to the circular path in the plane of the circular path.
  • Mathematically Tangential acceleration is written as

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and at = 3 m/s2

  • Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

Centripetal Acceleration (ac):

Hence, net acceleration 

A force vs extension graph of a spring is as shown. The work done in extending the spring is

  1. 10 J
  2. 5 J
  3. 5000 J
  4. 500 J

Answer (Detailed Solution Below)

Option 2 : 5 J

Kinematics and Kinetics Question 9 Detailed Solution

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Concept:

Work done = Area under force deflection diagram

Calculation:

Given:

Gradual loading

Force = 100 N, deflection = 100 mm = 0.1 m

A pin jointed uniform rigid rod of weight W and Length L is supported horizontally by an external force F as shown in the figure below. The force F is suddenly removed. At the instant of force removal, the magnitude of vertical reaction developed at the support is

  1. zero
  2. W/4
  3. W/2
  4. W

Answer (Detailed Solution Below)

Option 2 : W/4

Kinematics and Kinetics Question 10 Detailed Solution

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Explanation:

When the Force F is suddenly removed, then due to W, the rod is in rotating condition with angular acceleration 

Thus the equation of motion:

Also, the centre of the rod accelerates with linear acceleration a;

   

Work done by conservative force is equal to

  1. Decrease in potential energy
  2. Increase in kinetic energy
  3. Increase in potential energy
  4. Decrease in kinetic energy

Answer (Detailed Solution Below)

Option 1 : Decrease in potential energy

Kinematics and Kinetics Question 11 Detailed Solution

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Explanation:

Non-conservative forces:

  • non-conservative force is one for which work depends on the path taken.
  • Friction is an example of a non-conservative force that changes mechanical energy into thermal energy.
  • Work Wnc done by a non-conservative force changes the mechanical energy of a system.
  • In equation form, Wnc = ΔKE + ΔPE or, equivalently, KEi + PEi = Wnc + KEf + PEf.

Conservative forces:

  • conservative force is one, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. 
  • Example potential energygravitational force etc.
  • Work done by conservative force is equal to decrease in potential energy.

A particle of mass m is moving in a horizontal circle of radius r under the action of centripetal force expressed as , where C is constant. The total energy of the particle is:

Answer (Detailed Solution Below)

Option 2 :

Kinematics and Kinetics Question 12 Detailed Solution

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Concept:

A body moving in a horizontal circle has only kinetic energy stored in it.

Centripetal force = 

where m is mass, v is the velocity of particle, and r is the radius.

The kinetic energy of a body is given by:

Calculation:

Given:

Centripetal force = 

Centripetal force = 

 

Don't Confuse with the negative sign in the value of K.E as constant C must be negative here so that K.E become positive. 

The motion of a body in x-y plane is represented by x = 4 – 9t and y = t2 where x, y are in metre. Find the magnitude of its absolute velocity at t = 6 sec.

  1. 2.68 m/s
  2. 5.4 km/hr
  3. 10.77 m/s
  4. 15.0 m/s

Answer (Detailed Solution Below)

Option 4 : 15.0 m/s

Kinematics and Kinetics Question 13 Detailed Solution

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Concept:

Absolute velocity is given by:

vx = velocity component in x direction = dx/dt

vy = velocity component in y direction = dy/dt

Calculation:

At t = 6 sec

x = 4 – 9t

vx = dx/dt = -9

y = t2

vy = dy/dt = 2t = 2(6) = 12

For the point moving on a straight line which of following is true.

  1. Has no radial component of acceleration
  2. Has no tangential component of acceleration
  3. Has both tangential and radial component of acceleration 
  4. Has no radial and tangential component of acceleration

Answer (Detailed Solution Below)

Option 1 : Has no radial component of acceleration

Kinematics and Kinetics Question 14 Detailed Solution

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Explanation:

Acceleration:

  • The object under motion can undergo a change in its speed. The measure of the rate of change in its speed along with direction with respect to time is called acceleration.
  • The motion of the object can be linear or circular.

Linear acceleration:

  • The acceleration involved in linear motion is called linear acceleration.
  • Here the acceleration is only due to the change in speed and no acceleration due to change in direction, therefore no radial component of acceleration.

Circular acceleration:

  • The acceleration involved in a circular motion is called angular acceleration.
  • In a circular motion, the acceleration experienced by the body towards the centre is called the centripetal acceleration which can be resolved into two-component.
  • A radial component and a tangential component depending upon the type of motion.

Radial acceleration (ar):

  • The acceleration of the object along the radius, directed towards the centre is called radial acceleration.

Tangential acceleration (at):

  • The tangential component is defined as the component of angular acceleration tangential to the circular path.

To take the longest possible jump, an athlete should make an angle of _____

  1. 90 degree with the ground
  2. 60 degree with the ground
  3. 45 degree with the ground
  4. 30 degree with the ground

Answer (Detailed Solution Below)

Option 3 : 45 degree with the ground

Kinematics and Kinetics Question 15 Detailed Solution

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Concept:

Projectile motion: 

When a particle is projected obliquely near the earth's surface, it moves simultaneously in horizontal and vertical directions. The path of such a particle is called projectile and the motion is called projectile motion.

Range of projectile:

  • The horizontal range of a projectile is the distance along the horizontal plane it would travel, before reaching the same vertical position as it started from.

Formulae in projectile motion:

where u = projected speed, θ = angle at which an object is thrown from the ground and g = acceleration due to gravity = 9.8 m/s2.

Calculation:

Given:

Range of a Projectile motion is given by (R):

For horizontal distance to be maximum:

sin 2θ = 1

∴ sin 2θ = sin 90° 

∴ θ = 45°.

∴ to take the longest possible jump, an athlete should make an angle of 45° with the ground.

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