Kinetic Energy MCQ Quiz - Objective Question with Answer for Kinetic Energy - Download Free PDF

Calculation:

The bullet of mass m = 20 g = 0.02 kg first collides with a plate of mass m1 = 1 kg and then with a second plate of mass m2 = 2.98 kg.

Let u be the initial velocity of the bullet, and v be its velocity after passing through m1. Let v1 be the velocity of m1 after the bullet passes through it.

Apply conservation of linear momentum before and after first collision:

mu = mv + m1v1      (1)

Now, the bullet embeds into the second plate m2, so they move with common velocity v1 (since all move together after collision).

Apply conservation of linear momentum again before and after second collision:

mv + m1v1 = (m + m2)v1      (2)

Now eliminate v1 from equations (1) and (2):

v / u = (m + m2) / (m + m1 + m2)

= (0.02 + 2.98) / (0.02 + 1 + 2.98) = 3.00 / 4.00 = 0.75

The bullet retains 75% of its velocity. So, percentage loss in velocity = 25%

Answer: 25%

Calculation:

It is known that \(P=mv\) and \(E=\dfrac{1}{2}mv^2\),

\( P=\sqrt{2mE}\)

If kinetic energy are equal then \( P \propto \sqrt{m}\),

Hence, heavier body possesses large momentum,

As \( M_1< M_2 \)

Therefore

\( M_1V_1 < M_2V_2 \)

CONCEPT:

• Kinetic energy: The energy needed to move the body of mass m from one point to another with stated velocity v is called kinetic energy.

The Kinetic energy is given as:

K.E = ½ × m V²

Where K.E = Kinetic Energy

m = mass of the object

V = Velocity of an object

CALCULATION:

Given that, m = 22 kg, v = 5 m/s

∴ K.E = ½ × 22 × 5²

K.E = 275 J

Calculation:

If E is the translational kinetic energy, then the correct relation can be derived using the ideal gas law and the kinetic theory of gases.

From the kinetic theory of gases, the translational kinetic energy (E) is related to the pressure (P), volume (V), and temperature (T) by the equation:

For an ideal gas, the internal energy E is given by:

E = (3/2)RT = (3/2) PV

Thus, the correct relation that holds good is:

PV = (2/3) E

Correct Answer: Option 4 - PV = (2/3) E

CONCEPT:

Translational Kinetic Energy and Pressure-Volume Relation

Translational kinetic energy is the energy possessed by an object due to its motion from one place to another.

In thermodynamics, the relationship between the pressure (P), volume (V), and the translational kinetic energy (E) of an ideal gas can be derived from the ideal gas law.

EXPLANATION:

Let's examine the given options:

Option 1: PV = E

This is incorrect. The pressure-volume product is not equal to the translational kinetic energy.

Option 2: PV = 32" id="MathJax-Element-159-Frame" role="presentation" style="position: relative;" tabindex="0">3232" id="MathJax-Element-794-Frame" role="presentation" style="position: relative;" tabindex="0">3232 32 $\frac{3}{2}$ E

This is incorrect. While 32" id="MathJax-Element-160-Frame" role="presentation" style="position: relative;" tabindex="0">3232" id="MathJax-Element-795-Frame" role="presentation" style="position: relative;" tabindex="0">3232 32 $\frac{3}{2}$ E is related to the total kinetic energy per mole of gas, it is not the correct relation for PV.

Option 3: PV = 3E

This is incorrect. The factor of 3 is not applicable in this context.

Option 4: PV = 23" id="MathJax-Element-161-Frame" role="presentation" style="position: relative;" tabindex="0">2323" id="MathJax-Element-796-Frame" role="presentation" style="position: relative;" tabindex="0">2323 23 $\frac{2}{3}$ E

This is correct. The pressure-volume product (PV) of an ideal gas is directly proportional to the translational kinetic energy (E) with the relation PV = 23" id="MathJax-Element-162-Frame" role="presentation" style="position: relative;" tabindex="0">2323" id="MathJax-Element-797-Frame" role="presentation" style="position: relative;" tabindex="0">2323 23 $\frac{2}{3}$ E.

Therefore, the correct answer is option 4: PV = 23" id="MathJax-Element-163-Frame" role="presentation" style="position: relative;" tabindex="0">2323" id="MathJax-Element-798-Frame" role="presentation" style="position: relative;" tabindex="0">2323 23 $\frac{2}{3}$ E.

CONCEPT:

• Kinetic energy: The energy needed to move the body of mass m from one point to another with stated velocity v is called kinetic energy.

The Kinetic energy is given as:

K.E = ½ × m V²

Where K.E = Kinetic Energy

m = mass of the object

V = Velocity of an object

CALCULATION:

Given that, m = 22 kg, v = 5 m/s

∴ K.E = ½ × 22 × 5²

K.E = 275 J

CONCEPT:

• Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.

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

Where m = mass of the body and v = velocity of the body

CALCULATION:

Given that:

Initial speed (u) = 10 m/s

Final speed (v) = 20 m/s

Let the total mass is m.

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

The ratio of final KE to initial KE is given by:

\(Ratio = \frac{\frac{1}{2}mv^2}{\frac{1}{2}mu^2} = \frac{v^2}{u^2} =\frac{20^2}{10^2}=4:1\)

CONCEPT:

Kinetic energy

• The energy possessed by a body due to the virtue of its motion is called kinetic energy.

\(⇒ KE=\frac{1}{2}mv^{2}\)

Where KE = kinetic energy, m = mass and v = velocity

CALCULATION:

Given: Mass of the object (m) = 2000 g = 2 Kg,  and Kinetic energy (KE) = 100 J

• The kinetic energy of the body is

\(⇒ 100=\frac{1}{2}\times 2\times v^{2}\)

• The object must be moving at a speed of

\(\Rightarrow v^2=\frac{100 \times 2}{2}= 100\)

v = 10 m/s

The correct option is 1.

CONCEPT:

Kinetic energy (KE): 

The energy possessed by a body by virtue of its motion is called kinetic energy.

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

Where m = mass of the body and v = velocity of the body

Momentum (P):

The product of mass and velocity is called momentum. The SI unit of momentum is kg m/s.

Momentum (P) = Mass (m) × Velocity (v)

Force (F):

The interaction that after applying on a body changes or try to change the state of motion or state of rest is called force.

Force (F) = Mass (m) × acceleration (a)

The equation of motion is given by:

v = u + a t

Where v is final velocity, u is initial velocity, a is acceleration and t is time.

CALCULATION:

Given that:

⇒ Mass of the object (m) = 8 kg , Momentum (P) = 16 kg m/s, Applied force (F) = 0.4 N, Time (t) = 20 sec

⇒ We know that, Momentum (P) = Mass (m) × velocity (v)  

⇒ 16 = 8 × u 

⇒ u = 2 m/s = Initial velocity

⇒ Force (F) = Mass (m) × Acceleration (a)

0.4 = 8 × a

⇒ Acceleration (a) = 0.05 m/s²

⇒ Final velocity (v) = u + at = 2 + 0.05 × 20 = 3 m/s

⇒ So, change in kinetic energy = Final KE - Initial KE 

\(\frac{1}{2}mv^2 -\frac{1}{2}mu^2\)

\(=\frac{1}{2}\times8\times[v^2 -u^2]\)

\(=\frac{1}{2}.8.[3^2 -2^2]=20 \; J\)

⇒ Thus, the increase in the kinetic energy is 20 J.

The correct answer is reduce to one-fourth of its original value.

Important Points

• Kinetic Energy=  \(1 \over 2 \)\(mv^2\) (m is the mass of the body and v is the velocity at which the body is moving)
• So, when the velocity is reduced by half so the equation becomes
• Kinetic Energy = \(1 \over 2 \)\(mv^2\) = \({1\over 2} m \left( \frac {v}{2}\right)^2\) = \({1\over 2} m {v^2\over 4}\)  = \({1\over 4} \left({1\over 2} mv^2\right)\)
• From the above equation, we can see the kinetic energy is one-fourth the original value.

Key Points

• Kinetic Energy is the energy possessed by a body by the virtue of its motion.
• If a body is moving in a horizontal circle then its kinetic energy is the same at all points, but if it moving in a vertical circle, then the kinetic energy is different at different points.
• Example: A bullet fired from a gun can pierce a target due to its kinetic energy.

Additional Information

• The energy of an object is its capacity for doing work.
• Types of energy
  • Mechanical energy
  • Heat energy
  • Sound energy
  • Light energy
  • Chemical energy 
• Mechanical energy is of two types: Kinetic energy and Potential energy.
• Potential energy is the energy possessed by a body by virtue of its position.

CONCEPT:

• Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.

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

Where m = mass of the body and v = velocity of the body

EXPLANATION:

• A ball thrown upwards has some velocity so it has kinetic energy.
• A battery is always at rest. It doesn't possess kinetic energy. The battery has chemical energy inside it.
• A moving car has some velocity so it possesses kinetic energy.
• A flying bird also has some speed. So possess kinetic energy.

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.

Calculation:

Let initial speed is v, and Kinetic energy is K

Then

\(K = \frac{1}{2}mv^2\) -- (1)

Now, the speed is four times.

New speed v' = 4 v

the Kinetic energy will become 

\(K' = \frac{1}{2}mv'^2 = \frac{1}{2}m(4v)^2 =16\times \frac{1}{2}m(v)^2\)

\(\implies K' = 16 \times \frac{1}{2}m(v)^2 \) -- (2)

Putting (1) in (2)

K' = 16 K

So the kinetic energy becomes 16 times. 

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]

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

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

EXPLANATION:

So, from the given condition if kinetic energy is denoted as E and linear momentum is denoted as p by using the above relation, we can say that:

⇒ \(E = \frac{{{p^2}}}{{2m}}\) 

CONCEPT:

• Newton’s First Law: When a body is in state of rest then it will be in rest and when the body is moving then it will be moving until we apply an external force on it.
  • This property of resisting the change of state is called as inertia of the body.
• 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.

\(Force\;\left( F \right) = \frac{{{\bf{\Delta }}P}}{{{\bf{\Delta }}T}}\)

Where Δ P is Change in momentum and Δ t is change in time taken

ΔP = P₂ – P₁

Where P₂ is final momentum and P₁ is initial momentum of the system

• Newton’s third law of motion: For each action force there is an equal and opposite force.

EXPLANATION:

• From Newton's second law: Work on a free, rigid body, is equal to the change in kinetic energy of the velocity and rotation of that body,

W = ΔKE

• The work of forces generated by potential energy and the forces are said to be conservative.
• Therefore work on an object that is merely displaced in a conservative force field, without change in velocity or rotation, is equal to minus the change of potential energy of the object.

W = −ΔPE

These relations shows that work is the energy associated with the action of a force, hence the total work done on a particle is equal to the change in its kinetic energy always.

Concept:

Momentum

• Momentum is defined as the product of the mass and velocity of the body. 
• It is a vector quantity directed toward velocity. 
• It is given as 

p = mv -- (1)

m is mass, v is the velocity, p is momentum

Kinetic Energy

• The energy of a body in motion due to its state of motion is called kinetic energy. 
• It is given as 

\(K = \frac{1}{2}mv^2\) --- (2)

K is kinetic energy, m is mass, v is speed.

Relationship between Momentum and Kinetic Energy

If we combine equation (1) and (2) we will get the relationship between momentum and kinetic energy as

\(K = \frac{p^2}{2m}\) -- (3)

Calculation:

Given, kinetic energy is the same.

Let the mass of body 1 is m₁, body 2 is m₂, the momentum of body 1 is p₁ and body 2 is p₂.

Then from equation three we can say that 

\(K = \frac{p_{1}^2}{2m_1} = \frac{p_{2}^2}{2m_2} \)

Clearly, momentum is directly proportional to the mass if Kientic Enegy is constant. 

p = 2mK

K and 2 is constant.

P ∝ m

So, if mass will increase, momentum will increase. 

The heavy body will have greater momentum.