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Heavy water (D₂O) contains deuterium atom, i.e. hydrogen nuclei. They have nearly the same mass as those of neutrons. So, when neutrons strike against deuterium atoms, most of K.E. of the neutron is transferred to the deuterium atoms and thus neutrons get slowed down.

We know that

v_min = minimum speed needed at the lowest point, so that particle is just able to complete vertical circle = \(\sqrt{5gl}\)

Hence v_min = \(\sqrt{5\times 10\times 0.8}\) ms^-1

= 6.32 ms^-1

Also T₁ = Tension in the string at the lowest point of its circular path

= \(\frac{mv^2_1}{l}+mg\) = 6 mg

= 6 × 0.2 × 10 N

= 12 N

(i) When the body will slide down on an inclined plane then the frictional force will acts upwards to the inclined plane, so the angle between the motion of the body and the frictional force will be. 

Since, Work= FS cos θ 

F = force 

S = displacement 

θ = angle between direction of force and direction of motion = 180^o 

So, work done = F×s cos 180^o

Cos 180^o = -1 

So work done = - F×s 

Work done by friction on a body sliding down an inclined plane will be negative. 

(ii) In this case the man is doing work against the gravitational force, by applying force in an upward direction to lift the bucket. Since the direction of force applied and the direction of motion of bucket is the same, work done will be positive. 

Work= FS cos θ 

θ = angle between direction of force and direction of motion = 0^o 

Cos 0^o = 1 

So, work done = F × s

Fast blowing wind is used to run wind mills and also to generate electricity at small levels. Fast flowing stream is used to run turbines and ships.

Kinetic energy, K.E. = \(\frac{1}{2}m\vec{v}.\vec{v}=\frac{1}{2}m\vec{v}^2\)

Kinetic energy E = \(\frac{1}{2}\) (mv²)

m → mass of the body

v → velocity of the body.

No, because the work done is independent of time.

The Kinetic energy increases by four times.

Kinetic energy.

Freely falling body possess both potential & kinetic energy.

The energy possessed by a body due to its motion is called kinetic energy. A bullet fired from a gun can penetrate a target due to its kinetic energy.

Gravitational force.

A force is said to be non-conservative if the work done by the force depends on the path followed by the body, 

Example: Frictional force, Air resistance, Viscous force.

The winning team is performing wok over the losing team. Work done by winning team is positive while that of losing team is negative.

In physics no work is said to be done, if

(a) The applied force (F) is zero. A body moving with uniform velocity on a smooth surface has some displacement but no external force so in the case work done is zero.

(b) The displacement (S) is zero. A labourer standing with a load on his head doe no work.

(c)The angle between force and displacement (θ) is π/2 rad or 90° . Then cos θ = cos 90° = 0. Thus work done is also zero. In circular motion, instantaneous work done is always zero because of this reason.

(d) The change in kinetic energy (∆KE) is zero.

The ratio of relative velocity of separation after collision to the relative velocity of approach before collision.

Coefficient of restitution,

i. e., \(e = \frac{v_2-v_1}{u_1-u_2}\)

where u₁ and u₂ are initial velocities of the two colliding bodies and v₁, v₂ are their final velocities after collision.

(i) For elastic collision, velocity of separation is equal to the velocity of approach.

∴ e = 1

(ii) For inelastic collision, velocity of separation is not zero but always less than the velocity of approach.

∴ 0 < e < 1

(iii) For perfectly inelastic collision, the colliding bodies stick to each other and move with same velocity.

e = 0

The S. I unit of work is joule.

The energy possessed by a body due to its position or configuration is called potential energy. 

Example: The potential energy of water in dams is used to run turbines in order to produce electrical energy.

For maximum work θ = 0° and minimum work θ = 180°

Coefficient of restitution is defined as the ratio of relative velocity after collision to the relative velocity before collision. If u₁ & u₂ are the velocities of 2 bodies before collision & v₁ & v₂ are velocities after collision then coefficient of restitution e = \(\frac{v_1-v_2}{u_1-u_2}\) .

Consider a body of mass ‘m’ lying on the surface of earth. Force required to lift the body is equal to its weight mg. Let it be lifted through a height h. Work done on the body W = Fs = mgh This work done is stored in the body as its potential energy E_p = mgh.

Consider a body of mass ‘m’ at rest. Let a constant force F act on the body producing an acceleration ‘a’ for a distance ‘s’. Let the velocity of the body be changed to v. 

In this case, initial velocity = 0

final velocity = v.

Using, v² = u² + 2as

v² = 0 + 2as

v² = 2as

s =\(\frac {v^2}{2a}\)

Work done F × s

= m a s = ma \(\frac {v^2}{2a}\)

Work done = \(\frac {1}{2}\)mv²

By definition this work done is equal to the kinetic energy. 

∴ kinetic energy of the body, E_k = \(\frac {1}{2}\)mv².

it states that change in kinetic energy of a body is equal to work done and vice versa. Let a constant force \(\vec{F}\) be applied to a body moving with initial velocity \(\vec{u},\) so that its velocity becomes \(\vec{v}\) along the direction of force when S is its displacement. Using Newton’s second law of motion we get magnitude of force F = ma and from equation of motion, we get v² − u² = 2as, where a is the acceleration of the body.

Multiplying both sides by m/2, we get

\(\frac{1}{2}mv^2-\frac{1}{2}mu^2=maS\)

i.e., \(\frac{1}{2}mv^2-\frac{1}{2}mu^2=FS=W\)

i.e., K.E._(f) is final kinetic energy and K.E._(i) is initial kinetic energy.

Thus work done on a body by a net force is equal to the change in kinetic energy of the body.

The energy of a body is defined as its capacity for doing work.

Watt is the unit of power.

An elastic collision is the one in which both kinetic energy and momentum are conserved.

1. potential energy is the energy possessed by a body due to its position or configuration while kinetic energy is the energy possessed by a body due to its motion.

2. Gravitational potential energy is given by E_p = mgh, but kinetic energy is E_k = \(\frac {1}{2}\)mv².

3. Potential energy is measured by the amount of work that a body can do when it returns to the standard position. Kinetic energy is measured by the work done on the body to change its velocity.

4. A wound spring has potential energy while a bullet fired from a gun has kinetic energy.

Instantaneous power = Force × uniform velocity.

A force is said to be conservative if the work done by the force does not depend on path followed by the body but depends only on the initial and final positions of the body.

The relation between mass of a particle m and its equivalent energy (E) = mc² where, c = velocity of light in vacuum.

Work energy theorem states that work done by a force acting on a body is equal to the change produced in the K.E. of the body and vice versa.

The work-energy theorem states that the work done on a body is equal to the change in its kinetic energy.

The work energy theorem states that work done on a body is equal to the net change in its energy. (P.E or K.E)

Proof:

Consider a body of mass ‘m’ moving with an initial velocity u. Let a constant force F acting on a body changes its velocity to v. Let s be the distance traveled.

From the equation, v² = u² + 2as,

we get v² – u² = 2as

\(\frac {1}{2}\)(v² – u²) = as

Multiplying both sides by m,we have

\(\frac {1}{2}\)m(v² – u²) = mas

\(\frac {1}{2}\)m(v² – u²) = F.s

(∵ F = ma)

or \(\frac {1}{2}\)mv² – \(\frac {1}{2}\)mu² = W.

Although this theorem can be used to solve different types of problems in physics yet it does not give complete information about the real cause of motion (i.e., dynamics of Newton’s second law of motion).

It is called scalar form of Newton’s second law of motion.

Mass of ball A and B = m

Initial velocity of ball A = u₁ = v

Initial velocity of ball B = u₂ = 0

Final velocity of ball A = v₁ and of ball B = v₂

According to conservation of momentum

mv + m₂(0) = mv₁ + mv₂

mv = mv₁ + mv₂ …....(i)

Conservation of K.E.

\(\frac{1}{2}mv^2=\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2\)

\(mv^2=mv^2_1+mv^2_2\)

\(m(v^2-v^2_1)=mv^2_1\)

From (i) m(v - v₁) = mv₂ ......(ii)

Dividing equation (ii) by (i),

\(\frac{m(v+v_1)(v-v_1)}{m(v-v_1)}=\frac {mv^2_2}{mv_2}\)

or v₂ = v + v₁

Substituting in (i)

mv = mv₁ + m(v + v₁)

or v₁ = 0

Similarly solving v₂ = v

i. e., Ball A stops and ball B starts moving with velocity v, i. e., they exchange their velocities on collision.