Explore topic-wise InterviewSolutions in Current Affairs.

I = F × t = p₂ − p₁

Where notations have the usual meanings.

• No, it does not have kinetic energy if it does not have momentum. 
• Momentum is the product of mass and velocity. If it is zero, it implies that v = 0 (since mass can never be zero). 
• Now K.E = ~ mv², So if v = 0 then K.E also will be zero. 
• Thus, if an object has no momentum then it cannot possess kinetic energy.

The first law of Motion: Every object will remain at rest or in a state of uniform motion unless compelled to change its state by the action of a net force

C) pulling hands back while catching a ball

Flying a bird, swimming of a fish, launching a rocket, etc. are examples of Newton’s third law of motion.

Displacement is a scalar quantity.- False

B) mechanical energy

The second law of Motion: The rate of change of momentum of a body is directly proportional to the net force acting on it and it takes place in the direction of the net force.

The Law of conservation of momentum states in the absence of a net external force on the system, the momentum of the system remains unchanged.

Correct option is A) mass

Correct option is B) momentum

Inertia and linear momentum is measured by mass of the body and is a vector quantity and mass is a scalar quantity. 

The linear momentum of a body is the product of its mass and velocity. 

Momentum p = mass (m) × velocity (v) 

SI unit of momentum is Kg-m /sec or Ns.

Correct option is B) vector

Correct option is D) force

Linear momentum:

Momentum of a body is the product of its mass and velocity. Momentum p = m × v

a) F_net = 200 N ; ∆t =3 sec 

Impulse ∆p = F_net . ∆t = 200 × 3 = 600 N – sec.

b) Mass of the car (m) = 1200 kg.; Net force (F_net ) = 200 N

Time ∆t = 3 sec. ; Change in velocity ∆v =?

F_net = m. \(\frac{\Delta v}{\Delta t}\) 

200 = 1200 \(\times\,\frac{\Delta v}{3}\)

∆v = \(\frac{200}{400}= 0.5\,m/sec.\)

While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Correct option is A) 500 NS

Correct option is C) inertia

Correct option is C) velocity

A) Comes to rest

B) Newton’s laws of motion

Correct option is A) kg

Correct option is C) Impulse

Correct option is D) Mass

C) Newton’s third law of motion

C) The brick with greater mass moves less than the movement of the brick with lesser mass.

D) 120 cm/sec

(i) Let 

x₁, v₁ = position and velocity of the object at time t₁ respectively.

x₂, v₂ = position and velocity of the object at time t₂.

a = uniform acceleration of the object. 

v² - u² = 2as

Derivation – acceleration is given by

= \(\frac{v_2-v_1}{t_2-t_1}\),

where v₁ and v₂, t₁ and t₂ are as in …(i)

or t₂ − t₁ = \(\frac{v_2-v_1}{a}\) …(i)

Since, x₂ − x₁ = v₁ (t₂ − t₁ ) + \(\frac{1}{2}\) a(t₂ − t₁)² …(ii)

From (i) and (ii), we get

\(x_2-x_1=v_1\frac{v_2-v_1}{a}+\frac{1}{2}a{[\frac{v_2-v_1}{a}]}^2\)

= \(\frac{v_1v_2-v^2_1}{a}+\frac{v_2^2v_1^2-2v_1v_2}{2a}\)

= \(\frac{2v_1v_2-2v^2_1+v_1^2+v^2_2-2v_1v_2}{2a}\)

\(x_2-x_1=\frac{v_2^2-v_1^2}{2a}\)   ...(iii)

Or  \(v_2^2-v_1^2=2a(x_2-x_1)\)  ...(iv)

Now if v₁ = u at t₁ = 0

v₂ = v at t₂ = t   ...(v)

x₂ - x₁ = s

Then from (iv) and (v), we get

v^₂ - u^₂ = 2as  ...(vi)

(ii) v =u + at

Derivation – By definition of acceleration,

a = \(\frac{v_2-v_1}{t_2-t_1}\)

or  v₂ - v₁ =a( t₂ - t₁ )

or  v_{2 =} v₁ + a( t₂ - t₁ )  ...(i)

where v₁ and v₂ are the velocities of an object at times t₁ and t₂ respectively.

If v₁ = u (initial velocity of the object)

at t₁ = 0

v₂ = v(final velocity of the object)

at t₂ = t

Then (i) reduce to v = u + at

(iii) \(s=ut +\frac{1}{2}at^2\)

Derivation – Also let v_av = average velocity in t₂ − t₁ interval.

By definition

\(v_{av}=\frac{x_2-x_1}{t_2-t_1}\)

or  \(x_2-x_1=v_{av}(t_2-t_1)\)   ...(i)

Since, v_av = \(\frac{v_1+v_2}{2}\)  ...(ii)

∴ From eqns. (i) and (ii),

\(x_2-x_1=\frac{v_1+v_2}{2}(t_2-t_1)\)   ...(iii)

Also, we know that

\(v_2=v_1+a(t_2-t_1)\)   ...(iv)

∴ From eqns. (iii) and (iv),

\(x_2-x_1=\frac{1}{2}[v_1+v_2+a(t_2-t_1)](t_2-t_1)\)

= \(v_1(t_2-t_1)+\frac{1}{2}a(t_2-t_1)^2\)   ...(v)

Now if, x₁ =x₀ at t₁ = 0

x₂ =x and t₂ = t

v₁ = u at t₁ = 0

v₂ = v at t₂ = t

Eqn (v) reduces to \(x=ut+\frac{1}{2}at^2\).

The force of friction acting between my feet and ground is responsible for my deceleration.

Inertial frame of reference :

(a) A frame of reference which is at rest or which is moving with a uniform velocity along a straight line is called an inertial frame of reference.

(b) In inertial frame of reference Newton’s laws of motion holds good. 

(c) Ideally no inertial frame exist in universe. For practical purpose a frame of reference may be considered as inertial it it’s acceleration is negligible with respect to the acceleration of the object to be observed. 

Example : The lift at rest, lift moving (up or down) with constant velocity.

Non inertial frame of reference : 

(a) Accelerated frame of references are called non-inertial frame of reference. 

(b) Newton’s laws of motion are not applicable in non-inertial frame of reference.

Example : Car moving in uniform circular motion, lift which is moving upward or downward with some acceleration, plane which is taking off.

The weighting machine measures the reaction force given by the floor. So when lift is going 

(i) downwards the weight measured is lesser.

No. The acceleration of the particle can also be zero if the vector sum of all the forces is zero, i.e. no net force acts on the particle. 

As F = ma so F = -kx = ma

a = \(\frac{-kx}{m}\)

for x = 20 cm, ⇒ a = -10 m/s²

Initial momentum = 0 

Using conservation of linear momentum mv + MV = 0

V = \(\frac{-mv}{M}\) ⇒ V = 2.5 m/s

The gun recoils back when it is being fired to conserve the linear momentum.

Initial momentum of the gun & that of shell is zero as they are at rest.

Recoil velocity of the gun v₁ = – 3 × 10^-2ms^-1

Mass of the gun m₁ = 1000kg

Mass of the shell m₂ = 1 kg

velocity of the shell v₂ =?

From the equation

m₁ u₁ + m₂ u₂ = m₁v₁ + m₂ v₂

0 = 1000(- 3 × 10^-2 ) + 1 .v²

∴ v₂ = 30ms^-1

The gun target distance,

s = v₂ × t

= 30 × 8

= 240m.

As velocity is related to speed and direction, it can be changed by :

• Changing the speed while keeping the direction constant. 
• Changing direction while keeping speed constant.
• Changing both speed as well as direction of motion.

Given: 

mass of the cannon = 50 kg

recoil speed = 0.25 m/s 

To find: 

Momentum = ? 

Formula: 

Momentum = m x v 

Solution: 

Momentum = m x v 

= 500 x 0.25 

= 125 kg m/s

The momentum of cannon is 125 kg m/s

Mass (m) = 5 kg ; Initial velocity (u) = 10 m/s.; Time (t) = 15 s 

Final velocity (v) = 25 m/s. 

Acceleration a = \(\frac{v\,-\,u}{t}=\frac{25\,-\,10}{15}=\frac{15}{15}=1\,m/s^2.\)

Force applied on the object F = ma = 5 × 1=5 N.

If we slide or try to slide a body over a surface the motion is resisted by a bonding between the body and the surface. This resistance is represented by a single force and is called friction. 

The force of friction is parallel to the surface and opposite to the direction of intended motion.

Types of Friction :

(1) Static friction

(2) Limiting friction 

(3) Kinetic or dynamic friction

By observing the changes in shape, size, colour and chemical properties.

No, this change (Breaking of Barbie doll) cannot be reversed.

It is all depends on situations:
1. By erasing the drawing we can undo the change. For example, if we draw with the help of pencil, we can erase the drawing by using an eraser. Then, it is a reversible change.
2. If we draw using a pen, we cannot erase the drawing. Then this change cannot be reversed, so it is irreversible change.

No, it is a permanent change.

An act or process through which something becomes different.