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Test: Impulse and Momentum- 2 - EmSAT Achieve MCQ


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10 Questions MCQ Test - Test: Impulse and Momentum- 2

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Test: Impulse and Momentum- 2 - Question 1

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 struck with the second ball. Both the balls after impact will move with a velocity

Detailed Solution for Test: Impulse and Momentum- 2 - Question 1

Concept:

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.

Test: Impulse and Momentum- 2 - Question 2

Find the initial velocity, if S = 3t2- 2t + 3.

Detailed Solution for Test: Impulse and Momentum- 2 - Question 2

Concept:

Velocity = S/t    

 Here, S is displacement and t is time For instantaneous velocity,

it expressed as  dS/dt   

Explanation:

We have, S= 3t2- 2t + 3 

Hence, Instantaneous velocity =  dS/dt   = (6t - 2)

 initial velocity will be found at t = 0
Hence, V|initial| = (6 × 0 - 2) = -2

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Test: Impulse and Momentum- 2 - Question 3

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- 2 - Question 3

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

OG' = 0.22 + 0.152 = 0.25 m

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

(K.E + P.E)initial = (K.E + P.E.)final

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- 2 - Question 4

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- 2 - Question 4
  • 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 the Impulse Momentum equation and states that the impulse or force intensity is equal to the 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 the duration of the net force.
Test: Impulse and Momentum- 2 - Question 5

Impulse can be obtained from the-

Detailed Solution for Test: Impulse and Momentum- 2 - Question 5

Concept:

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

J = F  × dt = ΔP 

According to Newton's second law, the Force can be defined as a moment per time.

This is known as the Impulse momentum equation.

Linear impulse =  = change in momentum in that direction. 

Application:

(i) A cricket player lowers his hands while catching the ball: by doing so the time of impact increases and hence the effect of force decreases.

(ii) When a person falls from a certain height on the floor, he receives more injuries as compared to falling on a heap of sand. It is because of increasing the time of impact hence decreasing the impact of force.

Test: Impulse and Momentum- 2 - Question 6

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- 2 - Question 6

Concept:

  • 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:

Test: Impulse and Momentum- 2 - Question 7

On which of the fundamental principle a jet engine works?

Detailed Solution for Test: Impulse and Momentum- 2 - Question 7

A jet engine is a type of reaction engine discharging a fast-moving jet that generates thrust by jet propulsion.

  • It works on the principle of conservation of linear momentum.
  • The reaction principle accelerates a mass in one direction and as we know that from Newton's third law of motion it will experience a reaction and hence it results in thrust in the opposite direction.

Working of Jet engine:

  1. Air is drawn into the system through a set of valves, and fuel is sprayed into the incoming air.
  2. Combustion occurs and pressure is built up in the closed combustion region, closing the inlet valves and then accelerating the column of gas in the tailpipe outward.
  3. The escape of gases in exhaust permits the combustion gases to expand and the inertia of the out moving column of gases leaving the system lowers the pressure in the combustion chamber, allowing a fresh charge to enter through the inlet valve and repeat the cycle.
Test: Impulse and Momentum- 2 - Question 8

If light and a heavy body have an equal kinetic energy of translation, then ________.

Detailed Solution for Test: Impulse and Momentum- 2 - Question 8

∴ the speed of the smaller object is higher than that of bigger.

The momentum of the smaller object:

The lighter body will have smaller momentum.

Test: Impulse and Momentum- 2 - Question 9

During elastic and inelastic collision, ________ is conserved.

Detailed Solution for Test: Impulse and Momentum- 2 - Question 9
  • 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- 2 - Question 10

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- 2 - Question 10

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

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