Physics Definition And Proof Based Problems

### Definition And Proof Based Problems

Definition And Proof Based Problems
Q 3214223150

Define a uniform circular motion. For uniform circular motion, prove that : Linear velocity v = r omega.

Solution:

If the speed of the particle in circular path remains constant, the motion is uniform circular motion. We know, the arc length x covered with an angular displacement theta is x = r theta.
Differentiating,

 (dx)/(dt) = r (d theta)/(dt) \ \ ∵ r is constant, :. v = r omega.
Q 3234323252

Define the terms resultant or equivalent of two forces. Two forces F_1 and F_2 acting at an angle theta on a body simultaneously have a resultant F. Show that theta = cos^(-1) [ ( F^2 - F_1^2 - F_2^2 ) // 2 F_1 F_2 ]

Solution:

The resultant vector of two or more vectors is defined as that single vector which produces the same effect as is produced by individual vectors together. If F_1 and F_ 2 are the two forces the magnitude of the resultant F is given by,

F = sqrt ( F_1^2 + F_2^2 + 2 F_1 F_2 cos theta)

The angle theta between them is,

 theta = cos^(-1) ( (F^2 - F_1^2 - F_2^2)/(2 F_1 F_2) )
Q 3254323254

Define dot product of two vectors and give its geometrical interpretation.

Solution:

Dot product of two vectors vec a and vec b is defined
as, vec a . vec b = | vec a | | vec b | cos theta where theta is the angle
between vec a and vec b.

Let vec (O A) = vec a , vec (O B) = vec b

and angle A O B = theta

Then, OT = | vec b | cos theta

vec a . vec b = | vec a | | vec b | cos theta

 = OA xx OT

vec a . vec b = OA. Projection of vec b on vec a .
Q 3214667559

Prove that the path of one projectile as seen from another projectile is a straight line.

Solution:

The coordinates of one projectile as seen from another projectile are :

X= x_1 - x_2 = (u_1 cos theta_1 - u_2 cos theta_2 ) t

Y =y_1 - y_2 ,

= ( u_1 sin theta_1) t - 1/2 g t^2 - ( u_2 sin theta_2) t + 1/2 g t^2

= (u_1 sin theta_1 - u_2 sin theta_2 ) t

:. Y/X = ( (u_1 sin theta_1 - u_2 sin theta_2)t)/( (u_1 cos theta_1 - u_2 cos theta_2)t)

 = ( u_1 sin theta_1 - u_2 sin theta_2)/( u_1 cos theta_1 - u_2 cos theta_2)

 = m  (constant)

or, Y = m X

This equation represents straight line. Hence proved.
Q 3284823757

A projectile is fired at an angle theta with the horizontal.
(a) Show that its trajectory is a parabola.
(b) Obtain expression for
(i) the maximum height attained.
(ii) the time of its flight and
(iii) the horizontal range.
(c) At what value of theta is the horizontal range maximum?
(d) Prove that, for a given velocity of projection, the horizontal range is same for theta and (90° - theta).

Solution:

When a body is projected in the air in any direction, then the body is called a projectile.
(a) Suppose a body is projected with velocity u at an angle theta with the horizontal, P(x, y) is any point on its trajectory at time t.
Horizontal component of velocity is unaffected by gravity, but the vertical component (u sin theta) changes due to gravity

:. x = ( u cos theta) t.

 y = ( u sin theta ) t - 1/2 g t^2

 = u sin theta xx x/(u cos theta) - 1/2 g ( x/(u cos theta))^2

 y = x tan theta - (gx^2)/(2 u^2 cos^2 theta) ............(i)

It represents the equation of a parabola, hence the path followed by a projectile is a parabola.
(b) The greatest vertical distance attained by the projectile above the horizontal plane from the point of projection is called maximum height.
Maximum height, LN = H

(i) At maximum height
v = 0
:. u^2 - u_y^2 = -2 \ \ g H, where
u_y = u sin theta
or (u sin theta)^2 = 2 \ \ g H
or H = (u^2 sin^2 theta)/(2g)

(ii) At maximum height
v = 0
:. 0 = u sin theta - g t
or  t = ( u sin theta)/g

But time of flight,

T = 2t = (2 u sin theta)/g

(iii) When the body returns to the same horizontal level y = 0

:. 0 = x tan theta - (gx^2)/( 2 u^2 cos^2 theta)

or  x tan theta = (gx^2)/( 2 u^2 cos^2 theta)

or  x = ( 2 u^2 sin theta cos theta)/g = (u^2 sin 2 theta)/g

But coordinates of M are (R, 0). Putting x = R, we have

 R = ( u^2 sin 2 theta)/g

(c)  theta = 45^o

(d) When an object is projected with velocity u making an angle theta with horizontal direction.

 R_1 = ( u^2 sin 2 theta)/g ...........(i)

When an object is projected with u making an angle (90° - theta)

R_2 = ( u^2 sin 2 (90^o - theta))/g

 = u^2/g sin (180° - 2 theta)

= u^2/g sin 2 theta ..............(ii)

from (i) and (ii) R_1 = R_2

:. The horizontal range is same for two complementary angles.
Q 3254823754

Derive an expression for the acceleration of a body of mass 'm' moving with a uniform speed 'v' in a circular path of radius 'r'.

Solution:

Consider a body in a circular path of radius r, with a speed v. The velocity direction is tangential at any point in the path. The position vectors at A and B are represented by two sides of an isosceles triangle first.
The change in position vector is indicated by bar(AB) = Delta r. The velocity at A and B are along the tangents at these points and the change in velocity will complete an isosceles triangle of velocities.

bar MN = bar Delta v Since the triangles are similar,

(Delta v)/(Delta r) = v/r => Delta v = v/r . Delta r

Lt_(Delta t -> 0) (Delta v)/(Delta t) = Lt_(Delta t -> 0) v/r (Delta r)/(Delta t)

:. (dv)/(dt) = v/r . v => a = v^2/r
Q 3264812755

What are co-initial and collinear vectors ?

Solution:

Two vectors having the same initial point are called co-initial vectors. Two vectors which either act along the same line or along parallel lines are called collinear vectors.
Q 3254223154

Derive a relation for the time taken by a projectile to reach the highest point and the maximum height attained ?

Solution:

Consider a projectile projected at an angle theta to the horizontal with velocity u. The horizontal and vertical components initially with velocity are u cos theta and u sin theta respectively. Vertical velocity at highest point is zero, due to acceleration due to gravity acting vertically downwards.
Using  v = u + a t we have,
0 = u sin theta - g t

 => t = (u sin theta)/g

The time to reach topmost point t = (u sin theta)/g

Using v^2 = u^2 + 2 a s, we have

0 = u^2 sin^2 theta - 2 g \ \ h_(m a x)

 h_(m a x) = ( u^2 sin^2 theta)/(2 g)
Q 3214812759

Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

Solution:

The required angle of projection for max. horizontal range is 45°.

R_(max) = u^2/g

Max. height,

 H = (u^2 sin^2 theta)/(2g)

 = ( u^2 sin^2 45^o)/(2g) = u^2/(4g)

 (R_(max))/H = (u^2//g)/(u^2//4g) = 4

or  R_(max) = 4 H
Q 3284223157

State parallelogram law of vector addition. Show that resultant of two vectors A and B inclined at an angle theta is R = sqrt ( A^2 + B^2 + 2 AB cos theta) .

Solution:

Parallelogram law of vector addition. If two vectors vec A and vec B represent two adjacent sides of a parallelogram, the sum of the vectors is represented by the diagonal of the parallelogram.

Let vec A and vec B be two vectors at an angle theta between them. According to the law of parallelogram of vectors, the diagonal of the parallelogram indicates the sum of the other two sides/vectors vec A and vec B.

| vec (OQ) | = | vec A + vec B | = sqrt ( O T^2 + T Q^2)

| vec R | = sqrt ( (A + B + cos theta)^2 + ( B sin theta)^2)

 R = sqrt ( A^2 + B^2 + 2 AB cos theta)

The resultant R is at an angle alpha to vec A given by,

 alpha = tan^(-1) ( (B sin theta)/(A + B cos theta) )
Q 3214112959

Prove the following statement, "For Elevation which exceed or fall short of 45° by equal amount, the range is equal."

Solution:

At 45^o, the projectile has maximum range.

At (45° - theta )  the range is,

 R_1 = (u^2 sin [ 2 ( 45^o - theta) ])/g

 = ( u^2 sin (90^o - 2 theta) )/g

 = (u^2 cos 2 theta) /g

At (45° + theta) the range is,

 R^2 = ( u^2 sin [ 2 (45^o + theta) ] )/g

 = (u^2 sin (90^o + 2 theta) )/g

 = (u^2 cos 2 theta) /g

 R_1 = R_2