Center of Mass and Linear Momentum MCQ Quiz - Objective Question with Answer for Center of Mass and Linear Momentum - Download Free PDF

Correct option is: (1) 21 NS

v₁ = √(2gh₁)

= √(2 × 9.8 × 40)

= √784 = 28 m/s

v₂ = √(2gh₂) = √(2 × 9.8 × 10)

= √196 = 14 m/s

Impulse = Δp = m(v_f − v_i) = m(v₂ − (−v₁))

= (1/2)(14 − (−28))

Concept:

• The Centre of mass of a body is defined as a position defined relative to an object or system of objects.
• It is the average position of all the parts of the system, weighted according to their masses.
• In the case of rigid objects with uniform density, the center of mass is located at the center point of the body.
• For example - The Center of mass of a uniform disc, ring, etc. would be at its center. In some bodies or systems of bodies, the center of mass doesn't lie on the object.
• The Centre of gravity is defined as a point through which the force of gravity acts on an object or system.
• If the gravitational field is assumed to be uniform. The center of gravity is then in exactly the same position as the center of mass. But it is not necessary that the center of gravity coincides with the center of mass.

Explanation:

• The Center of mass may or may not coincide with the center of gravity.

Couple:

• A pair of forces of equal magnitude but acting in opposite directions with different lines of action is known as a couple or torque.
• A couple produces rotation without translation.

Examples:

1. When we open the lid of a bottle by turning it, our fingers are applying a couple to the lid.

Additional Information

Concept:

Impulse

• When a large force works on a body for a very small time interval, it is called impulsive force.
• An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such case, we measure the total effect of force.
• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of change in momentum.
• Impulse of a force is a measure of the total effect of force.
• Impulse is a vector quantity and its direction is same as that of force.
• SI unit of impulse is Newton-second or kg-m-s-1.

\(\vec I = \mathop \smallint \nolimits_{{t_1}}^{{t_2}} \overrightarrow {F\;} \cdot dt\)

Where, I = impulse, F = force and dt = very small time interval

Hence according to given explanation above

• Impulse, \(I=~\int \frac{dp}{dt}.dt⇒ i.e.,~I=\left( {{p}_{2}}-{{p}_{1}} \right)=\text{ }\!\!Δ\!\!\text{ }p\)
• therefore we can say that impulse is equals to change in momentum

Calculation:

Mass of the ball (m) = 150 g = 0.15 kg

Initial velocity (u) = 12 m/s

Final velocity (v) = 20 m/s

Time = 0.01 s

Initial Momentum p₁ = 0.15 × (-12) = -1.8 kg m/s

Final Momentum p₂ = 0.15 kg × 20 = 3 kg m/s

Impulse = F × T = change of momentum

⇒ F × T = p₂ - p₁ = (3 - (-1.8)) = 4.8 

⇒ F × 0.01 = 4.8

⇒ F = 480 N -s

Solution:

The system involves a freight car of mass  M  moving with a velocity  v , and sand falling onto it at the rate  \(\frac{dm}{dt}\) . Initially, the horizontal speed of the sand is  v = 0 . 

The momentum of the system at time  t  is given by:

\(P(t) = Mv + 0\)

At time \( t + \Delta t\) , when an additional mass  \(\Delta m \) has been added to the freight car:

\(P(t + \Delta t) = (M + \Delta m)v\)

The change in momentum is:

\(P(t + \Delta t) - P(t) = v \Delta m\)

Dividing by \( \Delta t \) and taking the limit as \( \Delta t \to 0\) , the required force to keep the freight car moving at speed  v  is:

Thus, the correct answer is:

\(a) F = v \frac{dm}{dt} \)

This force is required to maintain the velocity of the freight car as sand continuously falls onto it.

Concept:

• Kinetic Energy (K): The energy of the body in motion by virtue of motion is called kinetic energy.

\(K = \frac{1}{2}mv^2\)

m is the mass of the body, v is the speed of the body.

• Momentum (P): The product of the mass and velocity of a body in motion is called momentum of the body.

p = mv

• Relationship between Kinetic Energy and Momentum: Kinetic energy is the square of momentum divided by mass. 

\(K = \frac{p^2}{2m}\)

Calculation:

Let Kinetic energy and linear momentum of the body with mass m by K and P₁ respectively, then 

\(K = \frac{(P_1)^2}{4m}\)--- (1)

Another body with mass 2m has the same kinetic energy K. Let, it has linear momentum P₂. then

\(K = \frac{(P_2)^2}{m}\) --- (2)

From Equation (1) and (2) we have

\(\frac{(P_1)^2}{4m} = \frac{(P_2)^2}{m}\)

⇒ \(\frac{(P_1)^2}{(P_2)^2} = \frac{4m}{m}\)

⇒ \(\frac{(P_1)}{(P_2)} = \sqrt{\frac{4}{1}}\)

⇒ \(\frac{(P_1)}{(P_2)} = \frac{2}{1}\)

So, the ratio of P₁ : P₂ = 2 : 1

So, 2 : 1 is the correct option.

Concept:

Momentum:

momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.

• The unit of momentum (P) is kg m/s.
• Dimension: [MLT^-1]

Law of conservation of Momentum: 

A conservation law states that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.

P₁ = P₂

m₁ v₁ = m₂ v₂

Where, P₁ = initial momentum of system, P₂ = final momentum of system, m₁ = mass of first object, v₁ = velocity of first object, m₂ = mass of second object and v₂ = velocity of second object.

Calculation:

Given:  m₁ = 2 kg    m₂ = 3 kg     u₁ = 5 m/s        u₂ ​=  0 m/s

Let the common velocity of the combined body be V m/s

Mass of combined body      M = 2 + 3 = 5 kg

Applying conservation of momentum:          

m₁ v₁ + m₂ v₂ = M V

⇒ (2 × 5) + (3 × 0) = 5 V

⇒ 10 + 0 = 5 V

⇒ V = 2 m/s

Hence the combined velocity of both the spheres is 2 m/s.

Concept:

Second Law of Motion:

• The rate of change of momentum is directly proportional to the applied force. 

\(⇒ F = \dfrac{Δ p}{t}\)

⇒ F × t = Δ p

Where F is force, t is time, and Δ p is change in momentum. 

Calculation:

Given: Initial momentum (p_i) = 10 kg m s-1, Force  (F) = 30 N, and time (t) = 3 second

Let final momentum be p_f

• Change in momentum is

⇒ Δ p = F × t

⇒ Δ p = 30 N × 3 s = 90 N s = 90 kg m s^-1

• As we know the change in momentum is equal to the difference between the final momentum and initial momentum i.e., 

⇒ Δ p = pf - pi 

⇒ p_f = Δ p + p_i

⇒ p_f = 90 kg m s^-1 + 10 kg m s^-1 = 100 kg m s^-1

• So, the correct option is 100 kg m s-1

CONCEPT:

• Momentum (P): The product of mass and velocity is called momentum.
• The SI unit of momentum is kg m/s.
• Momentum (P) = Mass (m) × Velocity (v)

EXPLANATION:

Since, P = m v

Let the initial velocity of the body be v

Since the Mass of the body is constant 

So, P₁ = m v      ----(i)

According to the question

The new velocity of the body = 2v 

New Momentum (P₂) = m × 2v

⇒ P2 = 2mv        ----(ii)

On dividing (ii) by (i), we get 

\(\frac {P_2} {P_1} = \frac{2mv}{mv}\)

⇒ P₂ = 2 P₁

∴ The momentum will be doubled.

Key Points

• If the velocity of a body is doubled then its momentum doubles because velocity is directly proportional to momentum. So option 2 is correct.

The correct answer is 500 kg.

CONCEPT:

• The type of collision in which there is a loss of kinetic energy and after the collision both the colliding particles move together is called perfectly inelastic collision.
• In this type of collision the coefficient of restitution is equal to 0.
• Momentum of the system remains constant in this collision.

• Momentum before collision (P1) = Momentum after the collision (P2)

P1 = m1u1 + m2u2

P2 = m1v1 + m2v2

CALCULATION:

Here, speed of the car is 0 m/s

⇒ After collision speed is 8 m/s

According to "Principle of conservation of momentum"

⇒ So, 8 = (m₁v₁ + m₂v₂)/(m₁ + m₂) where m₁ is the mass of truck, v₁ is the velocity of truck, v₂ is the velocity of car and m₂ is the mass of car.

⇒ 8 = (2000 × 10 + m₂ × 0)/(2000 + m₂)

⇒ 16000 + 8 m₂ = 20000

CONCEPT:

• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

• Momentum Conservation: When in a system, there is no external force then the total momentum (P) of the system will be conserved.
• Kinetic energy: The energy in a body due to its motion, is known as kinetic energy.

The kinetic energy in terms of momentum is given by:

\(K=\frac{P^2}{2m}\)

where K is the kinetic energy of the body, P is the momentum of the body and m is the mass of the body.

CALCULATION:

\(K=\frac{P^2}{2M}\)

Let initial value of P = 100, m = 100

After increase of 30% P = 130

Initial kinetic energy \(K_i=\frac{P_i^2}{2m}=\frac{100^2}{2\times 100}=50\)

Final kinetic energy \(K_f=\frac{P_f^2}{2m}=\frac{130^2}{2\times 100}=169/2\)

% Increase in Kinetic energy = \(\frac{(K_f-K_i)}{K_i}\times100 = \frac{(\frac{169}{2}-50)}{50}\times100 = 69\)%

So the correct answer is option 3.

CONCEPT:

• In classical mechanics when an object is projected upward with some initial velocity, and as it moves upward its initial velocity keeps on decreasing because of gravitational acceleration since it acted downward.
• At the highest point, the final velocity will become zero and the time taken to reach that point can be calculated using a kinematic equation 

v = u + gt

where v = final velocity, u = initial velocity, g = acceleration due to gravity and t = time

Given,

Distance travelled (s) = 10 m,

Final velocity (v) = 0 m/s,

Acceleration due to gravity (g) = 9.8 m/s², 

Acceleration of the object, a = –9.8 m/s² (upward motion)

• The velocity with which the object was thrown upwards will be

⇒ v² = u² + 2as 

⇒ 0 = u² + 2 × (–9.8 m/s²) × 10 m

⇒ –u² = –2 × 9.8 × 10 m²/s²

⇒ u = 196 m²/s²

CONCEPT:

• Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.
  • Initial Velocity: The initial velocity can be given as x components and y components.

ux = u cosθ

uy = u sinθ

Where u stands for initial velocity magnitude and θ refers to projectile angle.

• Maximum Height: The maximum height is reached when vy = 0.

\({\rm{h}} = \frac{{{{\rm{u}}^2}{{\sin }^2}{\rm{θ }}}}{{2{\rm{g}}}}\;\)

Where h is the maximum height.

• Range: The range of the motion is fixed by the condition y = 0.

\(R = \frac{{{u^2}sin2θ }}{g}\)

Where R is the total distance covered by the projectile.

Kinetic energy (KE) = (1/2)m V²

Where m is mass and V is the velocity

CALCULATION:

Given that:

(Angle with vertical = 30°)

θ = 60° (with horizontal 90o - 30o = 60o )

ux = u cosθ = u cos 60° = (1/2) u

uy = u sinθ = u sin 60° =√3/2 / 2

Kinetic energy on the ground (KE) = (1/2)m u2 = K

At the top, velocity = V = u / 2

Kinetic energy at top (KE') = (1/2)m (u/2)2 = mu²/8 = K/4

So option 1 is correct.

CONCEPT:

• Momentum: Momentum is the product of the mass and the velocity of an object or particle. It is a vector quantity.
  • The standard unit of momentum magnitude is the kilogram-meter per second (kg·m/s).

P = m v

Where P = momentum, m = mass of the body, v = velocity of body.

• Conservation of Linear Momentum: Conservation of Linear Momentum states that a body in motion retains its total momentum (product of mass and vector velocity) unless an external force is applied to it.
• In mathematical term

Initial momentum = Final momentum

P₁ = P₂

We can say that

M₁V₁ + M₂V₂ = 0

Where M₁ = mass of projectile, V₁ = velocity of projectile, M₂ = mass of tank and V₂ = Unknown recoil speed of tank

CALCULATION:

Given that M₁ = 3 kg, V₁ = 24 m/s, M₂ = 0.4 tonnes = 400 kg

Substituting these values in the formula:

M₁V₁ + M₂V₂ = 0

(3 × 24) + (400 × V₂) = 0

V₂ = -0.18 m/s,

The tank recoils in the backward direction with a magnitude of 0.18 m/s.

So option 2 is correct.

Concept:

position of the centre of mass:

\({x_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + \ldots + {m_n}{x_n}}}{{{m_1} + {m_2} + \ldots + {m_n}}}\)

The velocity of the centre of mass:

\({V_{cm}} = \frac{{{m_1}{v_1} + {m_2}{v_2} + \ldots + {m_n}{v_n}}}{{{m_1} + {m_2} + \ldots + {m_n}}}\)

Acceleration of the centre of mass:

\({a_{cm}} = \frac{{{m_1}{a_1} + {m_2}{a_2} + \ldots + {m_n}{a_n}}}{{{m_1} + {m_2} + \ldots + {m_n}}}\)

Calculations:

\({\vec V_{cm}} = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\)

\(= \frac{{2 \times 20 - 4 \times 10}}{{2 + 4}}\) (Negative because both are moving in the opposite direction)

CONCEPT:

• 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:

\(KE = \frac{1}{2}m{v^2}\)

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,

 \(KE = \frac{1}{2}m{v^2}\)

Divide numerator and denominator by m, we get

\(KE = \frac{1}{2}\frac{{{m^2}{v^2}}}{m} = \frac{1}{2}\frac{{\;{{\left( {mv} \right)}^2}}}{m} = \frac{1}{2}\frac{{{p^2}}}{m}\;\) [p = mv]

\(\therefore KE = \frac{1}{2}\frac{{{p^2}}}{m}\;\)

\(p = \sqrt {2mKE} \)

CALCULATION:

Given that:

K.E1 = K.E2= K.E (let say)

m₁ = 1 kg and m₂ = 4 kg

The relation between the momentum and the kinetic energy is given by:

\(P = \sqrt {2m\;K.E}\)

But as K.E is the same

∴ \(P \propto \sqrt m \)

Or, \(\frac{{{P_1}}}{{{P_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}} = \sqrt {\frac{{{1}}}{{{4}}}} =1: 2 \)

So option 1 is correct.