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General
physics-

A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be


v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s

A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be

physics-General

v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s
General
physics-

An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
Horizontal distance A B equals v t subscript O B end subscript
equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
equals 3333.33 blank m equals 3.33 blank k m

An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

physics-General
From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
Horizontal distance A B equals v t subscript O B end subscript
equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
equals 3333.33 blank m equals 3.33 blank k m
General
physics-

A particle of mass m attracted with a string of length l is just revolving on the vertical circle without slacking of the string. If v subscript A end subscript comma v subscript B end subscript and v subscript D end subscript are speed at position A comma B and D then

At A comma blank v subscript A end subscript equals square root of g l end root
At B comma blank v subscript B end subscript equals square root of 5 g l end root
and at D comma blank v subscript D end subscript equals square root of 3 g l end root
Thus, v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript
Also, T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis
So, D comma theta equals 90 degree
therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g

A particle of mass m attracted with a string of length l is just revolving on the vertical circle without slacking of the string. If v subscript A end subscript comma v subscript B end subscript and v subscript D end subscript are speed at position A comma B and D then

physics-General
At A comma blank v subscript A end subscript equals square root of g l end root
At B comma blank v subscript B end subscript equals square root of 5 g l end root
and at D comma blank v subscript D end subscript equals square root of 3 g l end root
Thus, v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript
Also, T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis
So, D comma theta equals 90 degree
therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g
General
physics-

A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction

A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

physics-General
From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction
General
physics-

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

Work done W equals Area under F-xgraph with proper sign W equals Area of triangle A B C + Area of rectangle C D E F + Area of rectangle F G H I + Area of I J K L

W equals open square brackets fraction numerator 1 over denominator 2 end fraction blank cross times 6 cross times 10 close square brackets plus open square brackets 4 cross times left parenthesis negative 5 right parenthesis close square brackets plus open square brackets 4 cross times 5 close square brackets plus open square brackets 2 cross times open parentheses negative 5 close parentheses close square brackets
rightwards double arrow W equals 30 minus 20 plus 20 minus 10 equals 20 blank J ….(i)
According to work energy theorem
K subscript f end subscript minus K subscript i end subscript equals W rightwards double arrow open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript minus open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript equals W
open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript equals open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript plus W
equals 25 blank J plus 20 blank J equals 45 blank J [Using (i)]

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

physics-General
Work done W equals Area under F-xgraph with proper sign W equals Area of triangle A B C + Area of rectangle C D E F + Area of rectangle F G H I + Area of I J K L

W equals open square brackets fraction numerator 1 over denominator 2 end fraction blank cross times 6 cross times 10 close square brackets plus open square brackets 4 cross times left parenthesis negative 5 right parenthesis close square brackets plus open square brackets 4 cross times 5 close square brackets plus open square brackets 2 cross times open parentheses negative 5 close parentheses close square brackets
rightwards double arrow W equals 30 minus 20 plus 20 minus 10 equals 20 blank J ….(i)
According to work energy theorem
K subscript f end subscript minus K subscript i end subscript equals W rightwards double arrow open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript minus open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript equals W
open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript equals open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript plus W
equals 25 blank J plus 20 blank J equals 45 blank J [Using (i)]
General
physics-

A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

Range of the projectile on an inclined plane (down the plane) is,
R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction

A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

physics-General
Range of the projectile on an inclined plane (down the plane) is,
R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction
General
physics-

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

Vertical height equals h equals l cos invisible function application 30 degree
Loss of potential energy equals m g h

equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

physics-General
Vertical height equals h equals l cos invisible function application 30 degree
Loss of potential energy equals m g h

equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l
General
physics-

Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

The resultant of 5 N along O C and 5 N along O A is
R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
equals square root of 72N along O B
The resultant of square root of 72 N along O B and square root of 72 N along O G is
R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.

Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

physics-General
The resultant of 5 N along O C and 5 N along O A is
R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
equals square root of 72N along O B
The resultant of square root of 72 N along O B and square root of 72 N along O G is
R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.
General
maths-

P(θ) and D space open parentheses theta plus pi over 2 close parentheses are the pts. on the ellipse b x to the power of 2 end exponent plus a to the power of 2 end exponent y to the power of 2 end exponent equals a to the power of 2 end exponent b to the power of 2 end exponent then C P to the power of 2 end exponent plus C D to the power of 2 end exponent equals

P(θ) and D space open parentheses theta plus pi over 2 close parentheses are the pts. on the ellipse b x to the power of 2 end exponent plus a to the power of 2 end exponent y to the power of 2 end exponent equals a to the power of 2 end exponent b to the power of 2 end exponent then C P to the power of 2 end exponent plus C D to the power of 2 end exponent equals

maths-General
General
physics-

A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

By applying law of conservation of energy
m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root

A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

physics-General
By applying law of conservation of energy
m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root
General
physics-

A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
V equals 500 blank m divided by s

A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

physics-General
R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
V equals 500 blank m divided by s
General
physics-

A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

Work done = Area covered in between force displacement curve and displacement axis
= Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J

A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

physics-General
Work done = Area covered in between force displacement curve and displacement axis
= Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J
General
physics-

A particle P is sliding down a frictionless hemispherical bowl. It passes the point Aat t equals 0. At this instant of time, the horizontal component of its velocity v. A bead Q of the same mass as P is ejected from A to t equals 0 along the horizontal string A B (see figure) with the speed v. Friction between the bead and the string may be neglected. Let t subscript p end subscript and t subscript Q end subscript be the respective time taken by P and Q to reach the point B. Then

For particle P, motion between A and C will be an accelerated one while between C and B a retarded one. But in any case horizontal component of it’s velocity will be greater than or equal to v on the other hand in case of particle Q, it is always equal to v. Horizontal displacement of both the particles are equal, so t subscript P end subscript less than t subscript Q end subscript

A particle P is sliding down a frictionless hemispherical bowl. It passes the point Aat t equals 0. At this instant of time, the horizontal component of its velocity v. A bead Q of the same mass as P is ejected from A to t equals 0 along the horizontal string A B (see figure) with the speed v. Friction between the bead and the string may be neglected. Let t subscript p end subscript and t subscript Q end subscript be the respective time taken by P and Q to reach the point B. Then

physics-General
For particle P, motion between A and C will be an accelerated one while between C and B a retarded one. But in any case horizontal component of it’s velocity will be greater than or equal to v on the other hand in case of particle Q, it is always equal to v. Horizontal displacement of both the particles are equal, so t subscript P end subscript less than t subscript Q end subscript
General
maths-

If the chords of contact of P space open parentheses x subscript 1 comma y subscript 1 close parentheses and Q space open parentheses x subscript 2 comma y subscript 2 close parentheses w.r.t the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 are at right angle then fraction numerator x subscript 1 end subscript x subscript 2 end subscript over denominator y subscript 1 end subscript y subscript 2 end subscript end fraction equals

If the chords of contact of P space open parentheses x subscript 1 comma y subscript 1 close parentheses and Q space open parentheses x subscript 2 comma y subscript 2 close parentheses w.r.t the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 are at right angle then fraction numerator x subscript 1 end subscript x subscript 2 end subscript over denominator y subscript 1 end subscript y subscript 2 end subscript end fraction equals

maths-General
General
maths-

The area of an ellipse is 8π sq. units dist. between the foci is 4 square root of 3 then e=

The area of an ellipse is 8π sq. units dist. between the foci is 4 square root of 3 then e=

maths-General