Impact and Virtual Work MCQ Quiz in मल्याळम - Objective Question with Answer for Impact and Virtual Work - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

Law of conservation of momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

So from the law of conservation of momentum, we get m₁v₁ = m₂v₂

m₁ = Mass of 1^st object

m₂ = Mass of 2^nd object

v₁ = Velocity of 1^st object

v₂ = velocity of 2^nd object

From this, we can find the recoil velocity.

Let’s assume both bodies start from rest. So when the shot is fired, the bullet of mass m₁ gains a velocity of v₁ and that impacts the larger body m₂. Hence it will have recoil and will move backward.

So now the recoil velocity will be (v₂) = m₁v₁ / m₁

_{Important Points}

In this regard, Newton’s laws of motion must be known. There are three laws, which are as follows

• Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

It gives us an idea about Inertia.

• Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration
• For every action, there is an equal and opposite reaction.

Explanation:

• Virtual work is the work done by a real force acting through a virtual displacement i.e. Virtual work done by real forces.
• A virtual displacement is any displacement consistent with the constraints of the structure, i.e., that satisfy the boundary conditions at the supports.
• A virtual force is any system of forces in equilibrium.
• The principle of virtual work states, for bodies in equilibrium, for a small arbitrary displacement, the total work done by the system is zero.
• In this method, the system is displaced through a small amount about a reference point and the work done by all the forces about the reference point is summed to zero to find the unknown reactions if any.
• The principle of virtual work states that, for a body to be in equilibrium, the virtual work should be zero.​

Principle of Virtual Work

It states that if a small imaginary displacement is given to the system under equilibrium. Then the algebraic sum of work done by the force and moment is always equal to zero or partial derivatives of its total potential energy with respect to each of the independent variable is zero.

If P1, P2,……………..Pn are force and

δ1, δ2…………………….δn are the corresponding displacement

If M1, M2,……………….Mn are moments and

δθ1, δθ2………………….. δθn  are corresponding angular displacement

Then according to the principle of virtual work total work is done is zero.

P1δ1 + P2δ2 …………… + M1 δθ1 + M­2 δθ2 + ……………….. = 0

Explanation:

Let the value of moment = M

Given:

θ = 60°, L = 0.5 m, F = 80 N

From triangle law,

\(\tan \left( {90 - \theta } \right) = \frac{y}{L}\)

 y = L.cot θ

δ_y = L (-cosec² θ)⋅δθ

δ_y = -L⋅cosec²θ⋅δθ

“Principal of virtual work states that for a particle to be in equilibrium, the total work of the forces acting on the particle is zero for any virtual displacement of the particle”.

Considering upward direction of forces to be +ve.

M⋅δθ + F⋅δy = 0

M⋅δθ + F⋅(- L⋅cosec²θ⋅) δθ = 0

M = F⋅L⋅cosec²θ

M = 80 × 0.5 × (cosec 60°)²

∴ M = 53.33 N.m

Explanation:

Momentum Conservation Principle:

• For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.
• That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

\(m_1~×~u_1~+~m_2~×~u_2~=~(m_1~+~m_2)~×~V\)

m₁ and m₂ are the masses of the two cars, u₁ and u₂ are the initial velocities of the cars, V = Final velocity of the cars after coupling

Elastic collision:

• An elastic collision is a collision in which there is no net loss in kinetic energy in the system as a result of the collision.
• Both momentum and kinetic energy are conserved quantities in elastic collisions.

Calculation:

Given:

m1 = 15 Kg, m₂ = 12 Kg, u1 = 1.5 m/s, u₂ = 0.75 m/s

∵ \(m_1~×~u_1~+~m_2~×~u_2~=~(m_1~+~m_2)~×~V\)

⇒ (15 × 1.5) + 12 × (-0.75) = (15 + 12) × V

⇒ 22.5 - 9 = 27 × V

⇒ V = \(\frac{13.5}{27}\) = 0.5 m/s

The speed of both the cars immediately after the coupling = 0.5 m/s

Concept:

Conservation of angular momentum,

L₁ = L₂

∴ I₁ω₁ = I₂ω₂

Calculation:

Given ω1 = 6π rad/s, ω₂ = 5π rad/s;

After mass m is added at a distance r, the new moment of inertia will be

I₂ = I₁ + mr²

⇒ I₂ = I₁ + 20 × 10^-3 × (0.1)²

⇒ I₂ = I₁ + 2× 10^-4

From the conservation equation,

I₁ × 6π = (I₁ + 2 × 10^-4) × 5π

1.2 I₁ = I₁ + 2 × 10^-4

0.2 I₁ = 2 × 10^-4

I₁ = 1 × 10^-3

∴ I₁ = 1 gm-m²

Concept:

Power:

• The rate of work done by electric energy is called power. It is denoted by P.
• The SI unit of power is the watt (W).
• Watt is a small unit that's why kilowatt-hour is used as the unit for electrical energy.
• Since watt is the energy consumption rate (SI unit of power) of 1 J/s.

The energy lost by burning all of the fuel in 8 seconds, 

Required Energy = Total Energy - E_lost  

\(E_{req}={Power~of~engine} × {Time}\)

Calculation:

Given:

Power = 1250 kW, Time = 8 sec, Total Energy = 10500 kJ

E_req = 1250 × 8 = 10000 kJ

10000 = 10500 - E_lost

E_lost = 500 kJ

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

P_i = P_f

EXPLANATION:

• The Jet engine works on the phenomenon of Conservation of Linear Momentum.
• It produces a large volume of gases through the combustion of fuel, which is allowed to escape in backward direction through a jet.
• The backward rushing gas gains large momentum because of high velocity.
• It thus imparts an equal and opposite momentum to the jet engine.
• The jet engine, thus, moves forward with high speed. So option 1 is correct.

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 m is the mass of the body, P is the momentum 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.

CALCULATION:

• Since there is no external force in the system of the gun and the bullet, all forces are internal. So the momentum of the system will be conserved.
• According to the law of conservation of momentum, 
  • The initial momentum of the system is zero (as there is no motion initial). So the final momentum of the system will also be zero.
  • This implies that the recoil momentum of the gun must be equal and opposite to that of the bullet (proved below).

Pinitial = Pfinal 

0 = Pbullet + Pgun

Pgun = - Pbullet

• Here negative sign says the momentum of the gun is opposite to the momentum of the bullet, and the magnitude is the same.
• So, the recoil momentum of the gun must be equal and opposite to that of the bullet.

Calculation:

M = 10 kg, m = 1 kg

Momentum conservation:

mu = MV

\(\frac{V}{u}=\frac{m}{M}\)

The ratio of kinetic energies of bullet and gun is:

\(\frac{K.E_b}{K.E_g}=\frac{\frac{1}{2}mu^2}{\frac{1}{2}MV^2}=\frac{mu^2}{MV^2}=\frac{m}{M}(\frac{M}{m})^2=\frac{M}{m}=10\)

Concept:

In perfectly elastic collision, both the balls will interchange their velocity after impact, i.e. velocity of ball A after impact = 0

Coefficient of restitution, \({\bf{e}} = \frac{{{{\bf{V}}_2} - {{\bf{V}}_1}}}{{{{\bf{U}}_1} - {{\bf{U}}_2}}} \)

1 = before impact, 2 = after impact

For perfectly elastic body, e = 1

For perfectly plastic body, e = 0

Explanation:

Moment of a Couple:

A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance “d".

The moment of a couple is defined as
Mo = F.d (using a scalar analysis) or as
Mo = r × F (using a vector analysis).
Here r is any position vector from the line of action of F to the line of action of F

• The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F × d.
• Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body.
• Moments due to couples can be added together using the same rules as adding any vectors.