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Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

(b) frictional force

Force of friction = 0.5 N per quintal 

f = 0.5 x 2000 = 1000 N 

m = 2000 quintals = 2000 x 100 kg 

sin θ = \(\frac{1}{50}\) , a – 2 m/s² 

In moving up an inclined plane, force required against gravity 

mg sin θ = 39200 N 

And force required to produce acceleration = ma

= 2000 x 100 x 2 = 40,0000 N 

Total force required = 1000 + 39,200 + 40,0000 = 

440200 N.

Correct answer is (c) inertia

This is due to the inertia of direction.

As the friction between the tyres and road reduces on a rainy day.

F = 48 N,R = 45 x 9.8 = 441N

µ =\(\frac{F'}{R}\) = 0.22

Angle of friction θ = tan^-1 0.22 = 12°24′

Correct answer is (b) Ns

(c) µs or µk

No. force is required to move a body with constant velocity.

So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

(c) applied force

Correct answer is (d) no unit

(b) resistance to change its state

This is due to the inertia of direction.

Due to inertia of direction. 

(a) Inertial frame

(b) is constant

(b) non-inertial frame

(b) non inertial frame

Correct answer is (b) inertia

Due to inertia of direction.

When a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.

(b) Inertia of direction

(a) increases linearly

(a) first law of motion

• The property of an object at rest will try to remain at rest is called “static inertia”.
• For example, when the bus which is at rest begins to move suddenly, the person standing in the bus falls backward because of static inertia of the body. 
• Bus pulls the legs forward while body is in the rest position (static inertia).

(b) only in rotating frames

Correct answer is (d) no unit

(a) Inertia of motion

For initial velocity, u = 0

x =1/2 at²...(i)

If distance travelled in next 't' second be y, then

x + y = 1/2 a(t +t)² = 2at²....(ii)

Form equation (i) and (ii) we have

x/ x +y = 1/4 ⇒ 4x = x + y ⇒ y = 3x.

(c) ice and ice

(c) static friction is not zero and kinetic friction is zero

This is because a sudden start of the bus brings motion to the bus as well as to our feet in contact with the floor of the bus. But the rest of our body opposes this motion because of its inertia.

Let F force is applied by finger on a body of mass M,

Under Balanced condition,

F N

The normal reaction of the wall on the book. The minimum upward frictional force needed to ensure that the block does not fall is Mg. the friction force = μN.

Thus, minimum value of F = \(\frac{Mg}{μ}\)

(d) µ_s mg cos θ

If F is the force of the finger on the book, F = N, the normal reaction of the wall on the book. The minimum upward frictional force needed to ensure that the book does not fall is Mg. The frictional force = mN. Thus, minimum value of F = Mg/μ .

mg sin 30° = F

\(\frac{1}{2}\)mg = F

F = \(\frac{1}{2}\) × 100 N

= 50 N

For B (at rest), W = F = 50 N.

(b) angle of repose

Both are free falling. Hence, there is no acceleration of one w.r.t. another. Therefore, relative speed remains constant (=40 m/s ).

The weight W for which the system is in equilibrium, W = 50 N

(a) Acceleration of the football will always be vertical downward and is called acceleration due to gravity (g). 

(b) When football reaches the highest point it is momentarily at rest velocity, v = 0. As, it is continuously retarded by acceleration due to gravity (g).

Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as: 

R – mg = ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

R = mg 

= 70 × 10 = 700 N

Reading on the weighing scale = 700/g = 700/10 = 70 kg

Mass of the man, m = 70 kg

Acceleration, a = 5 m/s² downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a)

= 70 (10 – 5) = 70 × 5

= 350 N

Reading on the weighing scale = 350/g = 350/10 = 35 kg

Mass of the man, m = 70 kg

Acceleration, a = 5 m/s² upward

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

R = m(g + a)

= 70 (10 + 5) = 70 × 15

= 1050 N

Reading on the weighing scale = 1050/g = 1050/10 = 105 kg

When the lift moves freely under gravity, acceleration a = g

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a) 

= m(g – g) = 0

Reading on the weighing scale = 0/g = 0 kg

The man will be in a state of weightlessness.

Let R be the reading of the scale, in newtons. Effective downward acceleration = (50g - R)/50 = g

R = 5g = 50N. (The weighing scale will show 5 kg).

(b) v₀ = 0, a₀ ≠ 0, 

(d) v₀ ≠0,a₀≠ 0.

Explanation: 

Since the external force is nonzero, there will be nonzero acceleration. Hence (b) and (d). 

It may be at rest at time t, then (b) otherwise (d).

(b) ≤ R

Explanation: 

As large number of particles is situated at a distance R from the origin. If particles are uniformly distributed and make a circular boundary around the origin, then centre of mass will be at the origin. While if the particles are not uniformly distributed, then centre of mass will lie between particle and origin. This implies the distance between centre of mass and Origin is always less than equal to R.

(d) is equal to g.

Explanation: 

Since there is no force in the horizontal direction there will be no acceleration of CoM in the horizontal direction whatever be the direction, mass or speed of the two balls. Only force is the gravitational force hence acceleration of CoM will be acceleration due to gravity g.

(b) inside the square plate

Explanation: 

The mass of circular plate and the square can be assumed as concentrated at their centers. Hence CoM of the system will be on the line joining their centers. Since the mass of square plate will be more than the circular plate due to the greater area, hence combined CoM will be towards square plate from the middle i.e. inside the square plate.