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Test: Impulse and Momentum - Mechanical Engineering MCQ


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10 Questions MCQ Test - Test: Impulse and Momentum

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Test: Impulse and Momentum - Question 1

During elastic and inelastic collision, ________ is conserved.

Detailed Solution for Test: Impulse and Momentum - Question 1
  • Momentum is conserved in all collisions.
  • In elastic collision, kinetic energy is also conserved.
  • In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after the collision.

Perfectly elastic collision:
If law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:
The law of conservation of momentum holds good during a collision while that of kinetic energy is not.
Coefficient of restitution (e)

  • For perfectly elastic collision, e = 1
  • For inelastic collision, e < 1
  • For perfectly inelastic collision, e = 0
Test: Impulse and Momentum - Question 2

Two balls of equal mass and of Perfectly elastic material are lying on the floor. One of the balls with velocity V is made to strike the second ball. Both the balls after impact will move with a velocity

Detailed Solution for Test: Impulse and Momentum - Question 2

This is the perfectly inelastic collision/impact of the case of two bodies/balls as both the balls stick after the collision.
Perfect Inelastic Collision:

  • Two bodies move together with the same velocity.
  • The coefficient of Restitution will be 0.
  • Momentum is Conserved.
  • Kinetic Energy will not be Conserved.

​Calculation:
Given:
Let two balls A and B have mass mA, mB respectively, and their initial velocities are uA and uB. After the collision, they will move with the same velocity, vo.
Given that mass of both balls are same.
So  mA = mB = m
uA = V,  uB = 0
From the Concept of Momentum Conservation:
mAuA + mBuB = (m+m)vo
mV = 2mvo
vo = V/2
Both the balls after impact will move with velocity v/2.

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Test: Impulse and Momentum - Question 3

During Perfectly Elastic Collision between two bodies, which of the following quantities always remains conserved?

Detailed Solution for Test: Impulse and Momentum - Question 3

The correct answer is Both total kinetic energy and total linear momentum

  • A perfectly inelastic collision is a hypothetical collision in which two objects collide and energy is not wasted and momentum is conserved.
  • While Perfectly elastic collision happens then the object’s shape remains unchanged or not deformed.
  • Most of the collisions in daily life are inelastic in nature.

Important Points

Test: Impulse and Momentum - Question 4

Calculate the velocity of a body having a mass of 9 kg and linear momentum of 63 kg m/s.

Detailed Solution for Test: Impulse and Momentum - Question 4

Linear momentum:

  • It is the product of the mass and velocity of an object.
  • Like velocity, linear momentum is a vector quantity possessing a direction as well as magnitude.

P = mv
where P = momentum and v = velovity.
Calculation:
Given:
Mass = 9 kg, P = 63 kg-m/sec
P = mv
63 = 9 × v
v = 7 m/s

Test: Impulse and Momentum - Question 5

The kinetic energy of a body is stated to increase by 300 percent. The corresponding increase in the momentum of the body will be ____%

Detailed Solution for Test: Impulse and Momentum - Question 5
  • Kinetic energy (K.E): The energy possessed by a body by the virtue of its motion is called kinetic energy.

The expression for kinetic energy is given by:
KE = (1/2)mv2
Where m = mass of the body and v = velocity of the body

  • Momentum (p): The product of mass and velocity is called momentum.

Momentum (p) = mass (m) × velocity (v)
The relationship between the kinetic energy and Linear momentum is given by:
As we know,
KE = (1/2)mv2
Divide numerator and denominator by m, we get

Calculation:
Let initial Kinetic energy = KE1 = E
Given that:
Final kinetic energy (K.E2) = K.E1 + 300 % of KE1 = E + 3E = 4E
The relation between the momentum and the kinetic energy is given by:

Final momentum (P') will be:

Increase in momentum (ΔP) = P' - P = 2P - P = P
% Increase = 

Test: Impulse and Momentum - Question 6

A rigid homogeneous uniform block of mass 1 kg, height h = 0.4 m and width b = 0.3 m is pinned at one corner and placed upright in a uniform gravitational field (g = 9.81 m/s2), supported by a roller in the configuration shown in the figure. A short duration (impulsive) force F, producing an impulse IF is applied at a height of d = 0.3 m from the bottom as shown. Assume all joints to be frictionless. The minimum value of IF required to topple the block is

Detailed Solution for Test: Impulse and Momentum - Question 6

To topple the block we need an impulse force such that the weight will generate a clockwise moment about point O.
This will happen only when a minimum impulse that can rotate the block such that the Centre of gravity of the body will pass through the point O or beyond.

h = 0.4m, b = 0.3 m, m =1kg

h' = OG' - h/2 = 0.25 - 0.2 = 0.05 m
Due to force F, the block will topple about A
So By energy balance

ω2 = 2 × 12 × 9.81 × 0.05
ω = 3.43 rad/s
Now,
Angular impulse = change in angular momentum
IF × d = Io × (ωi - ωf)
IF × 0.3 = 1/12 × ( 3.43 - 0)
IF = 0.953 Ns

Test: Impulse and Momentum - Question 7

The rate of change of momentum is equal to the applied force and it takes place in the direction of the force is a statement of:

Detailed Solution for Test: Impulse and Momentum - Question 7

Newton’s Second law of motion: The rate of change of momentum of any object is directly proportional to the applied force on the body.

F × ΔT  = ΔP
Where Δ P = Change in momentum and Δ T = change in time taken

  • The above equation is known as Impulse Momentum equation and states that the impulse or force intensity is equal to change in momentum.
  • According to the impulse-momentum equation, the change in momentum of an object depends on both the net force acting on the object and duration of the net force.
Test: Impulse and Momentum - Question 8

A 1.0 kg ball drops vertically onto the floor with a speed of 25 m/s. It rebounds with an initial speed of 10 m/s. The impulse action on the ball during contact will be

Detailed Solution for Test: Impulse and Momentum - Question 8

Impulse (J): 

  • It is defined as the integral of force with respect to time.
  • It is also defined as a change in the linear moment (P) with respect to time.
  • It is a vector quantity.

J = F × dt = ΔP 
Calculation:
Given:
m = 1.0 kg, v1 = 25 m/s, v2 = 10 m/s
Impulse = change in momentum
J = ΔP
J = m(v1 - v2)
J = 1.0 × (25 - (-10))
= 1× (25) - 1 ×(-10) = 35 N - s 
∴ J = 35 N-s

Test: Impulse and Momentum - Question 9

If the momentum of a body increases from 10 units to 25 units in 5 sec, then the force acting on it is

Detailed Solution for Test: Impulse and Momentum - Question 9

The correct answer is 3 units.

  • Newton's 2nd law of motion says the rate change of momentum is called force.
  • The equation for the change in momentum is F ×  t = m ×  Δv
  • If the momentum of a body increases from 10 units to 25 units in 5 seconds.
  • Then the change in the momentum is 25 – 10 = 15 units in time 5 seconds.
  • So the Force extract from this is 15/5 = 3 Newton.

Key Points

  • Momentum is a vector quantity that is the product of the mass (weight) of a particle and its velocity (speed).
    • SI unit: kilogram meter per second (kg⋅m/s).
  • A spinning object has angular momentum.
  • An object travelling with a velocity has linear momentum.
  • A change in momentum may result from an acceleration or a force or an impulse.
Test: Impulse and Momentum - Question 10

Two inelastic spheres of masses 10 kg each move with velocities of 15 m/s and 5 m/s respectively in the same direction. The loss in kinetic energy when they collide

Detailed Solution for Test: Impulse and Momentum - Question 10
  • Momentum is conserved in all collisions.
  • In elastic collision, kinetic energy is also conserved.
  • In an inelastic collision, kinetic energy is not conserved. In a perfectly inelastic collision, objects stick together after the collision.


Conservation of momentum:
m1v1 + m2v2 = (m1 + m2) vf
Loss of kinetic energy:

Calculation:
Given:
m1 = m2 = 10 kg, v1 = 15 m/s, v2 = 5 m/s
Loss of kinetic energy:

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