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Newton’s Second Law of Motion:-

Momentum:-

Momentum of a body is defined as the amount of motion contained in a body.

Quantity of motion or the momentum of the body depends upon,

(a) mass of the body.

(b) velocity of the body.

Therefore momentum of a body of mass ‘m’ and velocity ‘v’ will be,

\vec{p} = m\vec{v}

Quantitative definition:-

Momentum of a body is equal to the product of its mass and velocity. Momentum is a vector quantity and possesses the direction of velocity.

Units:-

S.I:- kg m s-1

C.G.S:- g cm s-1

Momentum can be put into following two categories.

Dimension:-

[MLT-1]

(a) Non-relativistic momentum:-

According to classical physics (or non-relativistic physics) which is based upon the concepts of Newton’s laws of motion, mass of a body is considered to be a constant quantity, independent of the velocity of body. In that case momentum \vec{p} is given by,

\vec{p} = m\vec{v}.

Thus, momentum of a body is a linear function of its velocity.

(b) Relativistic momentum:-

In accordance to Einstein’s special theory of relativity, mass of a body depends upon the relative velocity ‘v’ of the body with respect to the observer. If ‘m0’ is the mass of body observed by an observer at rest with respect to body, its relativistic mass ‘m’ is given by,

m = \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{^{2}}}}}

Therefore, momentum of a body according to the concepts of theory of relativity is given by,

\vec{p} = \frac{m_{0}\vec{v}}{\sqrt{1-\frac{v^{2}}{c^{^{2}}}}}

Thus, relativistic momentum is not a linear function of v.

Frame of Reference:-

A system of co-ordinates whose axes can be suitably chosen is said to be a frame of reference. For location of a point ‘P’ we need three co-ordinate x, y and z. For complete identification of an event we must know ‘t’ also, i.e., the time of the occurrence. Hence an event in characterized by four co-ordinates (x,y,z,t). A reference frame describing an event in these four co-ordinates is known a space time frame.

Inertial and Non-Inertial Frame of Reference:-

(a) Inertial Frame:-

A frame of reference either at rest or moving with a uniform velocity (zero acceleration) is known as inertial frame. All the laws of physics hold good in such a frame.

An inertial frame is endowed with the following characteristics:

(i) All the fundamental laws of physics are valid in inertial frames.

(ii) All the fundamental laws of physics assume the same mathematical shape in all inertial frames.

(iii) They are isotopic with respect to mechanical and optical experiments

(b) Non-Inertial or Accelerated Frame:- It is a frame of reference which is either having a uniform linear acceleration or is being rotated with uniform speed.

Newton’s Second Law:-

The rate of change of momentum of a body is directly proportional to the impressed force and takes place in the direction of the force.

Newton’s first law provides a qualitative definition of the force while second law provides a quantitative definition of the force.

Let \vec{v} be the instantaneous velocity of the body. Momentum \vec{p} of the body is given by,

\vec{p} = m\vec{v}

According to second law,

\vec{F}∝ (rate of change of momentum)

Or,

\vec{F}\alpha \frac{d\vec{p}}{dt}

Or,

\vec{F}\alpha \frac{d(m\vec{v})}{dt}

Or,

\vec{F}\ =k \frac{d(m\vec{v})}{dt}

Here ‘k’ is the constant of proportionality. Mass ‘m’ of a body is considered to be a constant quantity.

\vec{F}\ =k m\frac{d(\vec{v})}{dt}

or,

\vec{F}\ =k m\vec{a}

The units of force are also selected that ‘k’ becomes one.

Thus, if a unit force is chosen to be the force which produces a unit acceleration in a unit mass,

i.e., F = 1, m = 1 and a = 1.

Then, k = 1

So, Newton’s second law can be written , in mathematical form, as

\vec{F}\ =m\vec{a}

i.e., Force = (mass) (acceleration)

This provides us a measure of the force.

Here, if F = 0 then we find a = 0. This reminds us of first law of motion. That is, if net external force is absent, then there will be no change in state of motion, that means its acceleration is zero.

Further we can extend second law of motion, (in fact its decomposition) to three mutually perpendicular directions as per our coordinate system.

If components in x, y and z direction are Fx, Fy & Fz respectively, the three acceleration produced when Fx, Fy & Fz act simultaneously) in the body are, Now,

three-forces

 If we add three forces then resultant is called net external force.

  Similarly,

 add-three-forces

 is called net acceleration produced in the body.

Unit of Force:-

S.I:- Newton [kg.m/sec2]

C.G.S:- Dyne [g.cm/sec2]

Dimension:-

[MLT-2]

Impulse:-

Impulse of a force is defined as the change in momentum produced by the force and it is equal to the product of force and the time for which it acts. Therefore, a large force acting for a short time to produce a finite change in momentum which is called impulse of this force and the force acted is called impulsive force or force of impulse.

 

 

 

According to Newton’s second law of motion,

\vec{F}\ =m\vec{a} =m\frac{(\vec{v}-\vec{u})}{t}

or, 

\vec{F}t = m\vec{v}-m\vec{u} = \vec{J}

So, Impulse of a force = change in momentum.

If the force acts for a small duration of time, the force is called impulsive force.

As force is a variable quantity, thus impulse will be,

J = \int_{t_{1}}^{t_{2}}F dt

The area under F - t curve gives the magnitude of impulse.

Impulse is a vector quantity and its direction is same as the direction of  vector-f .

Unit of Impulse:- The unit in S.I. system is kgm/sec or newton -second.

Dimension:- MLT1

 

Problem 1:-

The Sun yacht Diana, designed to negative in the solar system using the pressure of the sunlight, has a sail area of 3.1 km2 and a mass of 930 kg. Near Earth’s orbit, the sun could exert a radiation force of 29 N on its sail. (a) What acceleration would such a force impart to the craft? (b) A small acceleration can produce large effects if it acts steadily for a long enough time. Starting from rest then, how far would the craft have moved after 1 day under these conditions? (c)  What would then be its speed? (See “The Wind from the Sun,” a fascinating science fiction account by Arthur C.Clarke of a Sun yacht race.)

Solution:-

(a)

 Given Data:-

Mass of the yacht Diana, m = 930 kg

Force exerted by the sun light, F = 29 N

Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a).

So, F = ma        …… (1)

From equation (1), the acceleration (a) of the body would be,

a = F/m    …… (2)

Putting the value of m and a in equation (2), the acceleration such force impart to the craft would be,

a = F/m   

= 29 N /930 kg

= (3.1×10-2 N/kg) (1 kg. m/s2 /1 N)

= 3.1×10-2 m/s2     …… (3)

Thus acceleration such force impart to the craft would be, 3.1×10-2 m/s2.

(b)

Given Data:-

Time, t = 1 day

= (1day) (24 h/1 day) (60 min/1 h) (60 s/1 min)

= 86400 s

Initial velocity, vi = 0

Acceleration, a = 3.1×10-2 m/s2    

From equation of motion, we know that,

Distance travelled by the body (x) = vi + ½ at2

So, x = vit+ ½ at2     …… (4)

Putting the value of vi, a and t in equation (4), the distance travelled by the craft will be,

x = vit+ ½ at2    

= 0+½ (3.1×10-2 m/s2) (86400 s)2     (Since, a = 3.1×10-2 m/s2 and t = 86400 s)

 =1.1571×108     …… (5)

Rounding off to two significant figures, the distance will be 1.2×108 m.

Thus from the above observation we conclude that, the craft have moved1.2×108 m after 1 day under these conditions.

(c)

Given data:

Acceleration, a = 3.1×10-2 m/s2

Time, t = 86400 s

Acceleration of an object is equal to the velocity of the object divided by time.

a = v/t

So, v = at    ……(6)

Putting the value of a and t in equation (6), velocity would be,

     v = at   

        = (3.1×10-2 m/s2) (86400 s)

        = 2678.4 m/s   

Rounding off to two significant figures, speed will be 2700 m/s.

Thus from the above observation we conclude that, speed will be 2700 m/s.

Problem 2:-

A car travelling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65 cm (with respect to the road) while being brought to rest by an inflated air bag. What force (assumed constant) acts on the passenger’s upper torso, which has a mass of 39 kg? 

Concept:-

Force acting (F) on the body is equal to the mass of the body (m) times deceleration of the body (a).

F = ma   …… (1)

Solution:-

First we have to find out the deceleration (a) of the car.

If v0 is the initial speed of car and v is the final speed of the car, then the average speed (vav) of the car will be,

vav, = ½ (v+ v0)    …… (2)

To obtain the average speed (vav) while the car is decelerating, substitute 53 km/h for v0 and 0 m/s for v in the equation vav = ½ (v+ v0),

vav = ½ (v+ v0)   

       = ½ ((53 km/h)+ (0 m/s))

      = (½ ×53 km/h) (1,000 m/1 km) (1 h/60 min) (1 min/60 s)

      = 7.4 m/s          …… (3)

But average speed (vav) is equal to the rate of change of displacement (x).

vav = x/ t

 So, t = x/ vav     …… (4)

To obtain the time of deceleration t, substitute 0.65 m for x and 7.4 m/s for vav in the equation t = x/ vav,

t = x/ vav   

     = 0.65 m /7.4 m/s

    = 8.8×10-2 s       …… (5)

Deceleration (a) is equal to rate of change of velocity.

So, a = Δ v /t

         = ((0) - (53 km/h))/ 8.8×10-2 s

        = (-53 km/h)/ 8.8×10-2 s

        = ((-53 km/h) (1,000 m/1 km) (1 h/60 min) (1 min/60 s))/ 8.8×10-2 s

        = (-14.7 m/s)/ (8.8×10-2 s)

         = -1.7×102 m/s2      ……(6)

To obtain the force (F) acting on the passengers upper torso having mass 39 kg, substitute 39 kg for mass m and -1.7×102 m/s2 for deceleration a in the equation,  F = ma,

F = ma

        = (39 kg) (-1.7×102 m/s2)

        = -6630 kg. m/s2

       = -(6630 kg. m/s2) (1 N/1 kg. m/s2)

       = -6630 N         …… (7)

Rounding off to two significant figures, the magnitude of the force will be 6600 N.

Problem 3:-

Workers are loading equipment into a freight elevator at the top floor of a building. However, they overload the elevator and the worn cable snaps. The mass of the loaded elevatorat the time of the accident is 1600 kg. As the elevator falls, the guide rails exert a constant retarding force of 3700 N on the elevator. At what speed does the elevator hit the bottom of the shaft 72 m below?

Concept:-

Force acting on the body (F) is equal to the product of mas of the body (m) and acceleration of the body (a).

So, F = ma 

From equation F = ma, the acceleration (a) of the body would be,

a = F/m   

Weight W of the object is equal to the mass m of the object times of the free fall acceleration g.

W = mg

In accordance to equation of motion, the distance y travelled by the body will be,

y = ut + ½ at2   

Here u is the initial velocity, t is the time, and a is the acceleration.

When the elevator falls, the initial velocity u will be equal to zero.

So, u = 0

Substitute 0 for u in the equation y = ut + ½ at2,

y = ut + ½ at2

  = 0×t +½ at2

 = ½ at2

So the time t will be,

t = √2y/a

Speed v is equal to the product of acceleration a of the body and time t.

So, v = at

Solution:-

To obtain the weight of the elevator W, substitute 1600 kg for mass of the elevator m and 9.81 m/s2 for free fall acceleration g in the equation W =  (m) (g),

W =  (m) (g)

= (1600 kg) (9.81 m/s2)

   =15680 kg,m/s2

   = (15680 kg,m/s2) (1 N/1 kg,m/s2)

   = 15680 N

The magnitude of the net force F will be,

F = W-R

To obtain the magnitude of the net force F, substitute 15680 N for W and 3700 N for retarding force R in the equation F = W-R,

F = W-R

   = (15680 N) – (3700 N)

   =11980 N

Rounding off to two significant figures, the magnitude of the net force F will be 12000 N.

To obtain the acceleration a, substitute 12000 N for F and 1600 kg for m in the equation a = F/m, we get,

 a = F/m

    = 12000 N/1600 kg

    = (7.5 N/kg) (1 kg.m/s2/1 N)

    = 7.5 m/s2

To obtain the time t to fall, substitute -72 m for y and -7.5 m/s2 for a in the equation t = √2y/a,

t = √2y/a

  = √2(-72 m) /(-7.5 m/s2)

  = 4.4 s

To obtain the final speed v at which the elevator hits the bottom of the shaft 72 m below, substitute 7.5 m/s2 (only magnitude of a) for a and 4.4 s for t in the equation v = at, we get,

v = at

  = (7.5 m/s2) (4.4 s)

  = 33 m/s

From the above observation we conclude that, the speed at which the elevator hits the bottom of the shaft 72 m below would be 33 m/s.    

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