Messages posted by elessar_iitkgp

Let the initial speed be u and height reached be h.
For the upward journey,
0 = u + (F/m - g)T₁
T₁ = u / (g -F/m) ---------(1)

And

0 = u² +2(F/m - g)h
h = u² /2(g - F/m) ---(2)

For the downward journey,
-h = (1/2)[ -(F/m + g)]T_2²T₂ =

[(2u² /2(g - F/m)/)/(F/m + g)] (Substituting the value of h from (2))
T₂= u/

(g² - (F/m)²) -------(3)

Now,
(g + F/m)>(g - F/m)
Multiplying both sides by (g - F/m)
(g² - (F/m)²)>(g - F/m)²

(g² - (F/m)²)>(g - F/m)
u/

(g² - (F/m)²)<u / (g -F/m)
T₂<T₁

Thats what we are here for

Read HC Verma very well. Solve its problems. Then move on to MCQs by Bharati Bhawan. Want More?? Try your hands on Irodov

As the functions to be formed are onto, the two elements in the co-domain set must have at least one pre-image.
For the first of these two elements, the number of ways of selecting a pre-image is n & for the second element is n-1
No. of ways of selecting one distinct pre-image is n(n-1).
The rest n-2 elements in the domain set can link to either of the two elements in the co-domain set. Basically the problem reduces to dividing this set of n-2 elements into two sets, each containing atleast one element. The number of ways to do this is n-3.

Hence the total number of ways to do this is n(n-1)(n-3)

This formula is valid only for n>3.
For n=1, no onto functions are possible.
For n =2, only one onto function is possible.
For n= 3, the no. of ways of selecting one preimage for the two elements is 3x2 = 6. For the remaining one element in the domain there are only two choices. Hence the total number of functions is 6 x 2 = 12. But out of these only 6 are distinct.

Transform the integral as follows

I =

(

cos2x)/sinx =

(

(cos² x - sin² x))/sinx =

(

(1 - tan² x))/tanx
(Dividing the Nr & Dr by cosx)

Substitute tanx = t

I =

[(

(1-t² ))/t][1/(1+t² )] dt =

[(

(1-t² ))/t(1+t² )] dt

Substitute 1-t² = u²

I = -

[u /t(2-u² )][u/t} du = -

[u² /(1-u² )(2-u² )]du =

[u² /(u² -1)(2-u² )]du
=(1/2)

[2(u² -1) + (2-u² )/(u² -1)(2-u² )]du = (1/2)[

2/( (2-u² ) du -

1/(1-u²) du]
= (1/

2)ln[(

2 + u)/(

2 - u)] - (1/ 2)ln[(

2 + u)/(

2 - u)]
=(1/

2)ln[(

2 +

(1 - tan² x))/(

2 -

(1 - tan² x))] - (1/ 2)ln[(

2 +

(1 - tan² x))/(

2 -

(1 - tan² x))]
=(1/

2)ln[(

2 cosx +

(cos² x - sin² x))/(

2 cosx -

(cos² x - sin² x))] - (1/ 2)ln[(

2 cosx +

(cos² x - sin² x))/(

2 cosx -

(cos² x - sin² x))]
= (1/

2)ln[(

2 cosx +

(cos2x))/(

2 cosx -

(cos2x)]
- (1/ 2)ln[(

2 cosx +

(cos2x))/(

2 cosx -

(cos2x))]

Hope the answer si right ... Lengthy ...phew!!

Differentiation is essentially the process of finding the slope of the tangent of a given curve. When we say that a function is differentiable at a point, what we essentially mean is that its curve has a unique tangent at the point specified.

A non differentiable function doesn't have a unique tangent at the point under consideration. For example, consider the graph of |x|. At the origin you can have two tangents. So it isn't differentiable at x = 0. However for all other points it is differentiable.

The base is obviously greater than unity. Any base greater than 1 raised to a large power tend to infinity
So the answer is infinity

Agree with Nick ... the answer is 5h/3

Effective acceleration of the bodies
a_A = g+R/m_Aa_B = g+R/m_BHeight attained by the bodies
H_A = u² /2a_AH_B = u² /2a_B

Let m_A >m_Bg+R/m_A < g+R/m_Bu² /2a_A > u² /2a_BH_A >H_BHence the heavier mass attains greater height.

The rotation of earth around its own axis makes it a non inertial frame, for if we are to observe the equilibrium of any particle in the reference frame of earth, we mus introduce pseudo force. The effect of earth rotation is the reason for the coriolis forces that are observed on any particle moving on earth's surface, when observed in the reference frame of earth. Look up the last solved problem in HC Verma's Circular motion. You will get a fairly good idea of coriolis forces from there.

A similar effect is caused by the rotation of earth around the sun.

Now as to why we treat the earth as inertial frame for most problems. First these effects can be neglected for most problems that we solve at +2 level. Second, the complexity of the problem increases twofold.
So for a simpler analysis, we neglect such effects & assume that Newton's Laws are applicable.

Most of you have read projectiles and know how to find the equation of trajectory when the initial velocity, angle of projection are given.
Lets consider a reverse problem:
Given a general quadratic equation, lets find its initial velocity, angle of projection, time of flight, etc., etc, etc ....
Consider a quadratic equation
y = ax²+bx+c -------------(1)
The X axis is considered to be the horizontal plane, and the Y axis is vetial, positive up.
Now, for this equation to represent a projectile, it should be a parabola that opens downwards, ie, a<0
Also, it should have two real roots. For the case of a single real root, the equation represents a particle thrown vertically upwards.
So

= b²- 4ac >0
Lets assume that the particle is projected from the X axis only.
Then, if c>0, the particle is projected from the left of the origin.
If c = 0, the particle is projected from origin.
If c< 0, the particle is projected from the right of the origin.
Differentiating the equation wrt x,
dy/dx = 2ax + b
Equating it to zero, we get the critical point
x_c= -b/2a
d²y/dx²= 2a <0
Hence x_c is a point of maxima.
H = y_max = y(x_c) = -

/4a -----------------(2)
Differentiating the equation wrt time t,
dy/dt = 2ax(dx/dt) + b(dx/dt) ---------------(3)
Differentiating again wrt t
d²y/dt²= (2ax + b)(d²x/dt²) + 2a(dx/dt)² ---------------(4)
Now,
d²y/dt² = -g & d²x/dt² = 0
Substituting these values in (4)
-g = (2a)v_x²
v_x =

(-g/2a), which is a constant. -----------------(5)
So the X component of velocity remains constant.
From (3),
v_y= (2ax+b)v_x -------------------(6)
Let the roots of the equation (1) be

&

such that

<

. Then x =

is the projection point. Hence at the projection point,
u_x =

(-g/2a) &
u_y= (2a

+b)v_xNow as

is a root of the equation (1)

= (-b+

)/2a
(2a

+b) =

Hence at projection point,
u_x =

(-g/2a) &
u_y=

(-g

/2a)
So initial speed,
u =

u_x²+ u_y²=

(1+

)(-g/2a) --------------------(7)
Angle of projection

= tan^-1 (u_y / u_x) = tan^-1

------------------(8)
Time of flight
T = 2u_y/g =

(-

/2ag)
Horizontal range
R = 2u_y u_x/g = -

/a ------------------(9)
These derived equations provide a complete description of the projectile.
u =

[(1+

)(-g/2a)]

= tan^-1

R = -

/a
T =

(-

/2ag)
H = -

/4a
Lets take an example to see these formulas in work
Let the equation be
y = -x²+5x-6
The constant term is -ve, hence the particle is projected from the +ve X axis

= 1
u =

[(1+

)(-g/2a)] =

10 m/s

= tan^-1

= 45 deg
R = -

/a = 1 m
T =

(-

/2ag) = 1/(2

5) s
H = -

/4a = 1/4 m

y = 4/x
dy/dx = -4/x²
x(dy/dx) + xy²= - 4/x +(xy)y = -y + 4y = 3y

About potential energy ....

Lets take an example

Consider two masses kept at a distance from each other. Let one of the masses be fixed. The other mass experiences a gravitational attraction due to the fixed mass. If the free mass is not stopped, then it will accelerate towards the fixed mass and gain kinetic energy.

Now we will not let the free mass gain kinetic energy. Suppose an external agent applies a force equal and opposite to the gravitational force. Then the free mass is at equilibrium. Now suppose we move the free mass very slowly by an infinitesimal distance dx along the line joining the two particles, away from the fixed mass. The forces still approximately balance. (The gravitational force will change a little, but we may neglect it as the change is very small)

Now, the work done by the external force (against the gravitational force) during this small displacement does not increase the KE of the body, for the particle doesn't accelerate, So where does it go???

Now, work done always changes energy content. We define this work to correspond to the change in potential energy.
Hence
dW = F_ext dx = dU
As F_ext= F_gravdU = -F_grav dx

Or more generally,
dU =F_ext dx = -F_cons dx
That is change in PE is the amount of work done by an external force against a conservative force without changing the KE

Well, this is the question Einstein posed to his wife while explaining to her the basic difficulties of his theories.
For further details see any site that discuss Einsteins works.
However, the question is not in the realm of JEE

If d>f_con then parallel rays converge due to the lens at a point before they reach the mirror. In this case the focal length of the system is f_con .

If d<f_con
For parallel rays striking the lens they converge at focus. Hence for the mirror,
u = -(f_con- d) (A virtual object)
Applying the mirror formula,
1/v + 1/(f_con- d) = -1/f_div
v = - [(f_con- d) f_div ] /(f_con+ f_div-d)
Distance from the optic centre of the lens
[(f_con- d) f_div ] /(f_con+ f_div-d) + d
= (f_con f_div - (f_con+ d)d)/(f_con+ f_div-d)

Let the total time of motion be T and the height of the tower be H
Then distance fallen till T-1 time = H - (5/9)H = (4/9)H
So,
(4/9)H = (1/2)g(T-1)²----------(1)
H = (1/2)gT² ---------(2)

Substituting (2) in (1)
5 T² - 18T + 9 = 0
T = 3 and T = 3/5

But the stone falls for atleast one second as per the problem
So T = 3
Hence H = 45 m

Let the time interval between the drops be T. The 1st drop has traveled a for 4T time.
(1/2)g(4T)² = 32
T =

(2/5)

Distances travelled by the drops:
(1/2)g(T)², (1/2)g(2T)², (1/2)g(3T)²,32
2m, 8m, 18m, 32m

Distance between the drops
2m, 6m, 10m, 14m

Ok will correct the solution

Let the height of the building be h.
Let the topmost point of the building be A, The top of the window be B, the bottom of the window be C and the bottom most point of the building be D.

Height of the window, L = 1 m
Time taken by the ball to cross the window, T = 0.1 s
Time taken by the ball to travell from C to D and back to D again, T' = 2 s.
Acceleration, a = -g

Let the velocites of the ball at B & C be v_A & v_B respectively.
Then
v_B = v_A + aT

v_B = v_A - gT ----------------(1)

Also,
v_B² = v_A² + 2a(

y)

v_B² = v_A² + 2gL ---------------(2)

Solving equations (1) & (2) for v_B &v_A
v_B = -(gT + 2L/T)/2 = -10.5 m/s (Assuming g = 10m/s² )
v_A = (gT - 2L/T)/2 = -9.5 m/s

Let the distance from A to B be h₁, from C to D be h₂v_{A² =}2gh₁

h₁ = 4.5125 m

Now it can be shown that a ball dropped from a given height takes equal times to drop to the ground from that height and to come back to that height from the ground.

Hence time taken to travel from C to D
T' = 1 s

Hence,

y = v_B T' + (1/2)aT' ²

-h₂= v_B T' - (1/2)gT' ²

h₂= 15.5 m

Hence the height of the building = h₁ + L + h₂=21.0125 m

Let the point h below A be B an the point h above A be C
Taking upward direction positive,
v_C = +u (Say)
v_B = -2u
a = -g

y = -2h

Now applying the equation
v_B²=v_C²+ 2a(

y)
We get
u²= 4gh/3

Let velocity at A be u'
v_C²=v_A²+ 2a(

y')

v_A²= 10gh/3

Maximum height reached by the particle above A
H =v_A² /2g = 5h/3

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