Momentum Conservation MCQ Quiz in తెలుగు - Objective Question with Answer for Momentum Conservation - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

For perfectly elastic collision, both momentum and kinetic energy are conserved and the coefficient of restitution is equal to 1.

Calculation:

Given:

m₁ = 1 kg, u₁ = 12 m/s

m₂ = 2 kg, u₂ = 0 m/s

moment conservation: 

m1u1 + m2u2 = m1v1 + m2v2    

12 = v₁ + 2v₂     ........................ (1)

energy conservation:

\(\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_1}v_2^2\)

\(\frac{1}{2} \times 1 \times 144 + \frac{1}{2} \times 2 \times 0 = \frac{1}{2} \times 1 \times v_1^2 + \frac{1}{2} \times 2 \times v_2^2\)

\(⇒ 144 = v_1^2 + 2v_2^2\)    .................. (2) 

From (1) and (2)

\(144 = 144 + 4v_2^2 - 48{v_2} + 2v_2^2\)

\(6v_2^2 - 48{v_2} = 0\)

\(6{v_2}\left( {{v_2} - 8} \right) = 0\)

⇒v₂ = 8 m/s

Alternative method:

for perfectly elastic collision coefficient of restitution e = 1.

\(e = \frac{{velocity\;of\;seperation}}{{velocity\;of\;approach}}\)

for e = 1, velocity of seperation = velocity of approach

⇒ v₂ - v₁ = u₁ - u₂

⇒ v2 - v1 = 12  .................(3)

from (1) and (3)

3v₂ = 24

v₂ = 8 m/s

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 1st object

m₂ = Mass of 2nd object

v₁ = Velocity of 1st object

v₂ = velocity of 2nd object

 Concept:

In case of free fall of an object, the sum of kinetic energy + potential energy ie. Total energy at every point of the journey is constant.

\(i.e.\;{\left( {\frac{1}{2}m{V^2} + mgH} \right)_A} = {\left( {\frac{1}{2}m{V^2} + mgH} \right)_B}\)

Calculation:

Given mass m = 0.1 kg, height = 1 m

From,

\(\frac{1}{2}mV_A^2 + mg{H_A} = \frac{1}{2}mV_B^2 + mg{H_B}\)  (Between A & B)

\( \Rightarrow 0 + 0.1 \times 9.81 \times 1.0 = \frac{1}{2} \times 0.1 \times {V^2} + 0\)

\( \Rightarrow V = \sqrt {2 \times 9.81} \;\)

= 4.429 m/s

After impact, the velocity reduced by 20%

i.e. V₂ = 0.8 × V

= 0.8 × 4.429

= 3.543 m/s

Again from,

\(\frac{1}{2}mV_B^2 + mg{H_B} = \frac{1}{2}mV_c^2 + mg{H_c}\left( {between\;B\& C} \right)\)

\(\frac{1}{2} \times 0.1 \times {3.543^2} + 0 = 0 + 0.1 \times 9.81 \times h\)

\( \Rightarrow h = \;\frac{{{{3.543}^2}}}{{2 \times 9.81}}\)

= 0.639 m

≈ 0.64m

Concept:

\(e = \frac{{{V_2} - {V_1}}}{{{u_1} - {u_2}}}\)

Calculation:

For tennis boll, velocity of rebound is half the velocity of approach as bell takes twice the time to reach the initials position.

Let, suffix for ball is 1 and for wall is 2,

∴ u₂ = 0 and V₂ = 0

\({V_1} = \frac{{ - 1}}{2}{u_1}\) (Velocities are in opposite direction)

\(\therefore e = \frac{{{V_2} - {V_1}}}{{{u_1} - {u_2}\;}}\)

\(e = \frac{{0 - \left( {\frac{{ - 1}}{2}{u_1}} \right)}}{{{u_1} - 0}}\)

 ∴ e = 0.5

Concept:

• The product of mass and velocity is called the momentum of the body.
  Momentum (P) = m × V

Where,

m = mass of the body

V = velocity of the body.

Conservation of momentum:

• Whenever there is no net external force on the system then the total momentum of the system remains constant.

⇒ Initial momentum (P1) = Final momentum (P2)

Calculation:

Given: The cart A (m1) = 50 kg,u1 = 20 km/h, The cart B ( m2) = 20 kg,  u2 =10 kmph

• Two carts cling to each other so let the common velocity is v for the whole system.
• During the collision, the linear momentum is conserved hence, applying the conservation equation of momentum

⇒ 50 × 20 - 20 × 10 = 70 × v

⇒  \(v=\frac{80}{7}\) km/h

⇒  v = 11.43 km/h

Concept:

Conservation of momentum:

20 m = m v₁ + m v₂ ; v₁, v₂ → velocity of smooth and rough after impact

\(e = \frac{{velocity\;of\;separation}}{{velocity\;of\;approach}}\)

\(e = \frac{{{v_2} - {v_1}}}{{20}}\)

Perfectly elastic, e = 1

Calculation:

e = 1 ⇒ v₂ – v₁ = 20     

∴ v₂ = v₁ + 20

or  v₁ = v₂ – 20

Also, v₁ + v₂ = 20

∴ v₂ = 20 m/s, v₁ = 0 m/s

Friction retardation on rough block, a = μg

∴ Distance travelled = v² – u² = 2as

⇒ - 400 = - 2μgs

s = 200/μg

∴ s = 101.936 m

Note:

When perfect elastic impact occurs between two identical mass out of which one is at rest, then velocities are exchanged.

The rate of change of momentum is called force according to Newton’s second law of motion.

From the geometry, \(usin\alpha = vsin\beta\)    ………….. (1)

Coefficient of restitution is given by,

\(e = \frac{{0 - \left( { - vcos\beta} \right)}}{{ucos\alpha - 0}} \Rightarrow eu\;cos\alpha = v\;cos\beta\)   …………..(2)

And by dividing (2) by (1), we get

\(cot\beta = ecot\alpha\)

CONCEPT:

• Elastic collision: Elastic collision is a phenomenon where the collision of objects takes place such that the total linear momentum and kinetic energy of the system are conserved.
• Collisions in one dimension:

​Let m1 and m2 be the masses of two objects that undergo elastic collision. 

From the principle of momentum conservation,

⇒ m1v1i + m2v2i = m1v1f + m2v2f      

where m1, m2 are the masses of the colliding bodies, v1i, v2i are the initial velocity and v1f and v2f are their final velocities.

From the principle of kinetic energy conservation,

⇒ m1v1i2 + m2v2i2 = m1v1f2 + m2v2f2      

Calculation:

Velocity of the ball, just before making an impact with the incline

\(\rm u=\sqrt{2gh}=\sqrt{2\times9.8\times4.9}=9.8\) m/s

For perfectly elastic collision, along the line of impact,

velocity of approach = Velocity of separation

9.8 cos 30° = v cos θ    .......(i)

Now, momentum along the inclined plane will remain conserved,

⇒ 2u sin 30° = 2 v sin θ 

⇒ v sin θ = 9.8 sin 30° = 4.9

Using equation (i) and (ii), we get

v = \(\rm 9.8\sqrt{\sin^230+\cos^230^{\circ}}=9.8\) m/s

Therefore, θ = 30°

so, the inclination to the plane for the section = 90° - 30° = 60°

Now, momentum is given by,

\(\rm \vec P_f=2\times 9.8\sin60^{\circ}̂ i+2\times 9.8\cos 60^{\circ}\hat j\)

= 17 î + 9.8 ĵ

Concept:

• The conservation of momentum states that, within some problem domain, the amount of momentum remains constant; momentum is neither created nor destroyed, but only changed through the action of forces as described by Newton's laws of motion.
• Initial Momentum = Final momentum

Calculations:

Conserving momentum during the collision

mv(-j) + \({m\over 2}{v \over 2} \)(j) = (m + \({m \over 2}\))v_f 

-3 mv/4 j = 3m/2 v_f

v_f = v/2 (-j)

Before collision

\({GmM \over R^2 } = {mv^2 \over R}\)----(1)

v = \(\sqrt{Gm \over R} \)

After collision,

F_net = \({GMm \over R^2} -{m ({v \over 2})^2\over R}\)-----(2)

putting values of equation (1) in equation (2) , we get,

∴ F_net = \({3GMm \over 4R^2}\)towards the planet.

Since orbital velocity is v and after collision final velocity of the mass system is v/2​ which is less than orbital velocity,which implies that the mass will move in an elliptical orbit.

The correct answer is option (4).