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Test: Kinematics and Kinetics - Civil Engineering (CE) MCQ


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10 Questions MCQ Test Engineering Mechanics - Test: Kinematics and Kinetics

Test: Kinematics and Kinetics for Civil Engineering (CE) 2024 is part of Engineering Mechanics preparation. The Test: Kinematics and Kinetics questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Kinematics and Kinetics MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Kinematics and Kinetics below.
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Test: Kinematics and Kinetics - Question 1

A particle moves in a straight line. Its position is defined by the equation x = 6t2 − t3 where t in seconds and x is in meters. The maximum velocity of the particle during its motion will be

Detailed Solution for Test: Kinematics and Kinetics - Question 1

The velocity of the particle is defined as the rate of change of its displacement and can be expressed as
v = dx/dt
whereas, the speed of the particle at any instance is given as the rate of change of its distance.
for maximum velocity 
dv/dt = 0
Calculation:
Given :
Position: x= 6t2 - t3
dx/dt = V = 12t - 3t2   ...(i)
and for maximum velocity 
dv/dt = 0,
dv/dt = 12-6t = 0 or 6t = 12 or t = 2 sec
Substitute this value of t in the (i) 
we get v = 12× 2 - 3× 22 = 12 m/sec

Test: Kinematics and Kinetics - Question 2

The unit of linear acceleration is

Detailed Solution for Test: Kinematics and Kinetics - Question 2

Linear acceleration is defined as the rate of change of linear velocity of a body with respect to the time.
i.e a = v/t
and unit of velocity is m/s
so, unit of linear acceleration becomes m/s2.

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Test: Kinematics and Kinetics - Question 3

The angular velocity (in rad/s) of a body rotating at N r.p.m. is

Detailed Solution for Test: Kinematics and Kinetics - Question 3

Angular velocity is defined as the rate of change of angular displacement with respect to time. It is usually expressed by a Greek letter ω (omega).
Mathematically, angular velocity,
ω = dθ/dt
If a body is rotating at the rate of N r.p.m. (revolutions per minute), then its angular velocity,
ω = 2πΝ / 60 rad/s

Test: Kinematics and Kinetics - Question 4

A Particle is dropped from a height of 3 m on a horizantal floor, which has a coefficient of restitution with the ball of 1/2. The height to which the hall will rebound after striking the floor is

Detailed Solution for Test: Kinematics and Kinetics - Question 4

If a ball is dropped from a height ho on a horizontal floor, then it strikes with the floor with a speed,


and it rebounds from the floor with a speed, 

Where, Coefficient of restitution = 
First height of rebound, 


Therefore,

The velocity of ball after nth rebound; Vn = enVo
The height of ball  after nth rebound,  hn = e2nho
Calculation: 
Given: ho = 3 m., e = 0.5, h1 = ?
We know that, height after first rebound: 
h1 = (0.5)2 × 3 = 0.75 m.

Test: Kinematics and Kinetics - Question 5

During elastic impact, the relative velocity of the two bodies after impact is _______ the relative velocity of the two bodies before impact.

Detailed Solution for Test: Kinematics and Kinetics - Question 5

When two bodies impact, the Momentum is conserved, i.e.
m1u1 + m2u2 = m1v1 +  m2v2
Properties of different types of collision are given in the table below:

Perfectly elastic impact

  • Momentum is conserved, m1u1 + m2u2 = m1v1 +  m2v2
  • Kinetic Energy is conserved, 
  • The two bodies separate after the impact
  • There is no permanent deformation in the bodies during the impact.
  • In a perfectly elastic collision between equal masses of two bodies, velocities exchange on impact.
  • The relative velocity of the two bodies after impact is equal and opposite to the relative velocity of the two bodies before impact.

Plastic impact

  • The two bodies move together with a common velocity after the impact
  • Moment is conserved, i.e. m1u1 + m2u2 = (m1 + m2) v0
  • where v0 is the common velocity after impact.
  • There is a loss in Kinetic energy after the impact.
  • The coefficient of restitution is zero.
Test: Kinematics and Kinetics - Question 6

The linear velocity of a body rotating at ω rad/s along a circular path of radius r is given by

Detailed Solution for Test: Kinematics and Kinetics - Question 6

If the displacement is along a circular path, then the direction of linear velocity at any instant is along the tangent at that point.
therefore, the linear velocity will be ω.r

Test: Kinematics and Kinetics - Question 7

When a particle moves along a straight path, then the particle has

Detailed Solution for Test: Kinematics and Kinetics - Question 7

The acceleration of a particle at any instant moving along a circular path in a direction tangential to that instant, is known as tangential component of acceleration or tangential acceleration.

Test: Kinematics and Kinetics - Question 8

The co-efficient of restitution of a perfectly plastic impact is

Detailed Solution for Test: Kinematics and Kinetics - Question 8

Coefficient of restitution (e):
Coefficient of restitution or coefficient of the resilience of a collision is defined as the ratio of the relative velocity of separation after the collision to the relative velocity of the approach before the collision.


A perfectly inelastic collision also called a perfectly plastic collision is a limiting case of inelastic collision in which the two bodies stick together after impact.
Since both bodies stick together in a perfectly plastic collision. Therefore vA = vB, Thus e = 0
Properties of different types of collision are given in the table below:

Test: Kinematics and Kinetics - Question 9

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

Detailed Solution for Test: Kinematics and Kinetics - Question 9


When the Force F is suddenly removed, then due to W, the rod is in rotating condition with angular acceleration α
Thus the equation of motion:

Test: Kinematics and Kinetics - Question 10

When a particle moves with a uniform velocity along a circular path, then the particle has

Detailed Solution for Test: Kinematics and Kinetics - Question 10

The acceleration of a particle at any instant moving along a circular path in a direction normal to the tangent at that instant and directed towards the centre of the circular path is known as normal component of the acceleration or normal acceleration. It is also called radial or centripetal acceleration.

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