Physics 4 Marks Question

Explanation:

When the earth is closest to the sun, speed of the earth is maximum, hence KE is maximum. When the earth is farthest from the sun, speed is minimum, hence KE is minimum but never zero and negative.

2. (c) directly proportional to time

Explanation:

$\text{F}=\frac{\text{k}}{\text{v}}$

$\text{m}\frac{\text{dv}}{\text{dt}}=\frac{\text{K}}{\text{v}}$

$\Rightarrow\int\text{mv dv}=\int\text{K}\text{ dt}$

$\Rightarrow\text{m}\frac{\text{v}^2}{2}=\text{Kt}$

$\Rightarrow\text{Ke } \infty\text{ t}$

3. (a) 6.2 meV

Explanation:

The kinetic energy of an air molecule is

$\frac{10^{-21}\text{J}}{1.6\times10^{-19}\frac{\text{J}}{\text{eV}}}=0.0062\text{ eV}$

This is the same as 6.2 meV.

4. (c) 1 : 2

Explanation:

As we know that, linear momentum, p

$=\sqrt{2\text{mK}}$

$\bigg(\because\text{K}=\frac{\text{p}^2}{2\text{m}}\bigg)$

For same kinetic energy, $\text{P}\infty\sqrt{\text{m}}$

$\frac{\text{P}_1}{\text{P}_2}=\sqrt\frac{\text{m}_1}{\text{m}_2}=\sqrt\frac{1}{4}=\frac{1}{2}=1:2$

5. (d) 300%

Explanation:

Initial velocity = 10 ms^-1

Final velocity = 20 ms^-1

^{Final KE $=\frac{1}{2}\times10\times20\times20=20\times10^2\text{J}$}

% incrase $=\frac{(20-5)\times10^2}{5\times10^2}\times100=300\%$

1. (a) Commutative law

2. (a) Positive

3. The scalar product or dot product of any two vectors A and B, denoted as A.B (read A dot B) is defined as

$\text{A}\cdot\text{B}=\text{AB}\cos\text{q}.$ where q is the angle between the two vectors. Since A, B and $\cos\theta$ are scalars, the dot product of A and B is a scalar quantity.

Each vector, A and B, has a direction but their scalar product does not have a direction. Following are properties of dot product

the scalar product follows the commutative law :

A.B = B.A

Scalar product obeys the distributive law:

(B + C) = A.B + A.C

Further, A. $(\lambda\text{B})=\lambda(\text{A}\cdot\text{B})$ where $\lambda$ is a real number.

For unit vectors i, j, k we have

i × i = j × j = k × k = 1 and i × j = j × k = k × i = 0

• $\text{A}\times\text{A}=\mid\text{A}\parallel\text{A}\mid\cos\theta=\text{A}^2.$
• B = 0, if A and B are perpendicular.

4. The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement. Thus

$\text{W}=(\text{F}\cos\theta)\text{d}=\text{F}\cdot\text{d}$

Work has only magnitude and no direction. Its SI unit is (N m) or joule (J).

5. No work is done if:

The displacement is zero.

The force is zero. A block moving on a smooth horizontal table is not acted upon by a Horizontal force (since there is no friction), but may undergo a large displacement.

The force and displacement are mutually perpendicular. This is so since, for $\theta=\frac{\pi}{2}$rad $\cos\big(\frac{\pi}{2}\big)=\theta.$

1. (a) Scalar quantity

2. (d) All of the above

3. Work energy theorem: The change in kinetic energy of a particle is equal to the work done on it by the net force. Mathematically

K_f - K_i = W

Where Ki  and Kf are respectively the initial and final kinetic energies of the object. Work refers to the force and the displacement over which it acts. Work is done by a force on the body over a certain displacement. Energy possessed by object due to its motion is called as kinetic energy. Its SI unit is N-m or Joule (J).

4. Kinetic energy is scalar quantity as it is a work done and work done is scalar quantity hence kinetic energy is also scalar quantity and doesn’t have any direction.

5. The kinetic energy possessed by an object of mass, m and moving with a uniform velocity, v is

$\text{K}=\frac{1}{2}\times\text{mv}^2=\frac{1}{2}\text{v}.\text{v}$

Kinetic energy is a scalar quantity. Having unit the same as that of work, that is, joule (J).

1. (a) zero

Explanation:

When earth is moving around the sun in a circular orbit, then gravitational attraction on earth due to the sun provides required centripetal force, which is in radially inward direction, i. e. in a direction perpendicular to the direction of motion of the earth in its circular orbit around the sun. As a result, the work done on the earth by the force will be zero. i.e. W Fd = cos 90^°

2. (d) All of the above

Explanation:

Work done by weight-lifter is zero, because there is no displacement. In a locomotive, work done is zero because force due to gravity and displacement are mutually perpendicular to each other. In case of a person holding a suitcase on his head and standing at a bus terminal, work done is zero because there is no displacement.

3. (b) cos ^-1 (0.32)

Explanation:

Given, $\text{F}=(3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}})\text{unit.}$ 

and $\text{d}=(5\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}})\text{unit.}$

$\therefore\text{ f.d}=\text{F}_\text{x}\text{d}_\text{x}+\text{f}_\text{y}\text{d}_\text{y}+\text{f}_\text{z}\text{d}_\text{z}$

= 9 + 16 + 25

= 50 units

$\Rightarrow\text{d}=\sqrt{50} \text{units}$

$\text{ d.d}=\text{d}^2=\text{d}^2_\text{x}+\text{d}^2_\text{y}+\text{d}^2_\text{z}$

= 25 + 16 + 9

= 50 units.

$\Rightarrow\text{d}=\sqrt{50}\text{units}$

$\Rightarrow\cos\theta=\frac{16}{\sqrt{50}\sqrt{50}}$

$=\frac{16}{50}=0.32\big(\because\cos\theta=\frac{\text{F.d}}{\text{FD}}\big)$

4. (d) I, II and III

Explanation:

The work done in displacing an object by applying force F is given by W = F . s = F s cos $\theta$ So, work done will be zero, when

(i) either applied force F or displacement s is zero.
(ii) the force and displacement are mutually perpendicular to each other. i.e. $\theta$ = 90^°

5. (c) more for the case of a positron, as the positron moves away a larger distance

Explanation:

Force between two protons is same as that of between proton and a positron. As positron is much lighter than proton, it moves away through much larger distance compared to proton. We know that, work done = force × distance. As, forces are same in case of proton and positron but distance moved by positron is larger, hence work done will be more in case of positron.

1. (a) [ML²T^-2]

2. (c) Both a and b

3. The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground by height h to that point against gravity. Let the work done on the object against gravity be W. That is, work done,

W = force × displacement

= mg × h

Therefore potential energy (PE) = mg*h. The dimensions of potential energy are [ML²T^-2] and the unit is joule (J), the same as kinetic energy or work.

4. Let us consider some of the definitions of a conservative force.

• A force F(x) is conservative if it can be derived from a scalar quantity V(x).
• The work done by the conservative force depends only on the end points. This can be seen from the relation, W = Kf – Ki = V (xi ) – V(xf ) which depends on the end points.
• A third definition states that the work done by this force in a closed path is zero. This is once again apparent since xi = xf.

5. Conservation of mechanical energy:

Suppose that a body undergoes displacement Δx under the action of a conservative force F. Then from the WE theorem we have,

$ \triangle\text{K}=\text{F}(\text{x})\triangle\text{x}.$ If the force is conservative, the potential energy function V(x) can be defined such that $\triangle\text{v}=−\text{F}(\text{x})\triangle\text{x}.$ The above equations imply that

$\triangle\text{K}+\triangle\text{V}=0$ or $\triangle(\text{K}+\text{V})=0.$

Which means that K + V, Over the whole path, xi to xf, this means that

Ki + V(xi ) = Kf + V(xf ). The quantity K +V(x), is called the total mechanical energy of the system. Individually the kinetic energy K and the potential energy V(x) may vary from point to point, but the sum is a constant.

1. (b) If work is done against conservative force.

Explanation:

Potential energy of a body increases when work is done against a conservative force, e.g. if we raise the height of an object, its potential energy increases because work is done against gravitational force which is a conservative force.

2. (d) All of the above

Explanation:

The zero of the potential energy is arbitrary. It is set according to convenience. For the spring force, we took U (x), = 0, at x = 0, i.e. the unstretched spring had zero potential energy. For the constant gravitational force mg, we took U = 0 on the earth’s surface. Also, for the force due to the universal law of gravitation, the zero is best defined at an infinite distance from the gravitational source.

3. (b) 3 : 2

Explanation:

F = k₁ x₁ 1, F = k₂ x₂

_{$\text{K}_1\text{x}_1=\text{K}_2\text{x}_2\Rightarrow\frac{\text{K}_1}{\text{K}_2}=\frac{\text{x}_1}{\text{x}_2}$}

_{$\frac{\text{PE} \ (1)}{\text{PE}\ (2)}=\frac{\text{k}_1 \text{x}^2_1}{\text{k}_2\text{ x}^2_2}$}

_{$=\frac{\text{k}_1}{\text{k}_2}\times\Big(\frac{\text{k}_2}{\text{k}_1}\Big)=\frac{\text{k}_2}{\text{k}_1}=\frac{3}{2}$}

4. (a) 27 J

Explanation:

PE of spring $=\frac{1}{2}\text{kx}^2\Rightarrow\text{PE} \infty\text{x}^2$

$\therefore\text{PE}=15\times\frac{(4)^2}{(3)^2}=15\times\frac{16}{9}\simeq27\text{ J}$

5. (c) 30 J

Explanation:

Potential energy of the spring when stretched through a distance x,

$\text{U}=\frac{1}{2}\text{kx}^2=10\text{ J}$

When x becomes 2x, the potential energy will be

$\text{U}=\frac{1}{2}\text{k}(2\text{x})^2=4\times\frac{1}{2}\text{kx}^2$

$= 4 × 10 = 40 \text{ J}$

$\therefore\text{Work done}=\text{U}'-\text{U}=40-10=30\text{ J}$

1. (a) chemical energy

Explanation:

One of the greatest technical achievements of human kind occurred when we discovered how to ignite and control fire. We learnt to rub two flint stones together (mechanical energy), got them to heat up and to ignite a heap of dry leaves (chemical energy), which then provided sustained warmth.

Explanation:

Parabolic plots of the potential energy U and kinetic energy K of a block attached to a spring obey in a Hooke’s law. The two plots are complementary, one decreasing as the other increases. The total mechanical energy E = K + U remains constant.

3. (a) kinetic energy atB must be mgh:

Explanation:

In a conservative field loss of PE or gain of
KE depends only on initial and final point
and not on path covered, i.e. at B, KE = mgh.

4. (a) U > E

Explanation:

We know that, PE + KE = Mechanical energy

U + K = E
⇒ U = E - K
Now, K can never be negative, so
U < E
K = E - U

Now, U can be negative, so K > E is possible.

5. (d) 5m

Explanation:

According to the law of conservation of energy,

$\frac{1}{2}\text{mu}^2=\frac{1}{2}\big(\frac{1}{2}\text{mu}^2\big)+\text{mgh}$

$\Rightarrow490=245+5\times9.8\times\text{h}$

$\text{h}=\frac{245}{49}=5\text{m}$