physics. session kinematics - 5 session opener how far can this cannon fire the shot? courtesy :

Physics

Session

Kinematics - 5

Session Opener

How far can this cannon fire the shot?

Courtesy : http://physics.nad.ru/Physics/English/thermo.htm

Session Objectives

Session Objectives

1. Projectile motion (Definition)

2. Parameters of Projectile Motion

3. Path of Projectile Motion

4. Instantaneous radius of Curvature

5. Projectile motion on an inclined plane

Projectile motion (Definition)

Projectile Motion = Uniform horizontal motion + Non-Uniform vertical motion(constant acceleration due to gravity)

Assumptions :

(1) g=constant (y<<Rearth)(2) v<<vescape

(3) Air friction & force of buoyancy are neglected(4) Object is a particle

Projectile Motion

Courtesy : www.physicsclassroom.com

Gravity free path:

Vertical free fall:

Vx=constant

Vy= f(t)

Projectile motion

g=constant

Y

X

vtx=vcosvtY= vsin -gt

H

R

v

Parameters of Projectile Motion

2v sin2R

g

2v sinT

g

2 2v sinH

2g

R

x

y

90H

Path of Projectile Motion

22 2

gy x tan x

2v cos

t

2

v v gt

1s v t gt

2

333333333333333333333333333333333333333333

33333333333333333333333333 33

y

x

v v sin gt

v v cos (const.)

21y v sin t gt

2x v cos t

Class Exercise

Class Exercise - 1A stone is thrown with a velocity of 19.6 m/s at an angle 30° above horizontal from the top of a building 14.7 m high. Find

(i) the time after which the stone strikes the ground

(ii) the distance of the landing point of the stone from the building

(iii) the velocity with which the stone hits the ground

(iv) the maximum height attained by the stone above the ground

Solution

14.7 m

V

vx(0) = v cos = 19.6 cos30º = 9.8 3 m/s

vy(0) = v sin = 19.6 sin30º = 9.8 m/s

ax = 0, ay = -9.8 ms–2

(i) y(t) = vy(0)t + ayt21

2

–14.7 = 9.8t – 9.8t212

or 4.9t2 – 9.8t – 14.7 = 0

or t2 – 2t – 3 = 0

t = 3 s

Solution

(ii) R = vxt = = 50.92 m 9.8 3 3

vx(t) = 9.8 3 m/s

vy(t) = vy(0) + ayt

= 9.8 – 9.8 × 3

or vy = –19.6 m/s

V

Vx

Vy

2 2x yv v v 9.8 7 m/ s

y1

x

vtan

v

1 2

tan below horizontal3

Solution

Maximum height attained above the ground = Height of B above the point of projection + Height of building

2 2v sinH 14.7 14.7

2g

2 219.6 114.7 19.6 m

2 9.8 2

Class Exercise - 2

A projectile of mass m is projected with a speed v at an angle with the horizontal. Which one of the following graphs represents the variation of kinetic energy of the projectile with time?

KE

t

(a)

KE

t

(b) KE

t

(c)

KE

t

(d)

Solution

KE is least at the highest point as u sin = 0

Hence answer is (d)

Class Exercise - 3

A particle is projected upwards with a velocity of 120 m/s at an angle of 60° to the horizontal. The time after which the particle will be moving upwards at an angle of 45° to the horizontal will be(Take g = 10 m/s2)

(a) 6 3 1 (b) 6 3

(c) 3 3 1 (d) 6 3 1

Solution

120 cos60° = v cos45°

120 1

v2 2

v 60 2

v sin45 120 sin60 – gt

3

60 120 – gt2

120 3 – 110t

2t 6( 3 – 1)

Hence answer is (a)

Class Exercise - 5

If is the equation of a

trajectory, then the time of flight will be

21y x x

2

2 3(a) (b)

g g

4 2 2(c) (d)

g g

SolutionComparing the equation with that of the trajectory of a projectile.

Hence answer is (a)

2

2 2gx

y x tan2u cos

21y x x

2

tan = 1 or = 45° and u = 2g

12 2g

2usin45 22T

g g g

Class Exercise - 6A projectile is thrown at an angle = 30° with a velocity of 50 m/s in magnitude. Find the time after which the velocity of the projectile makes an angle of 60° with the initial velocity. (Take g = 10 m/s2)

Angle between u and v = 2 (Angle of projection) The particle will be at the same horizontal level finally.

22 50(Y )2usin

T 5 sg 10

Solution :

30° 30°30°

Class Exercise - 7

If u 80 i + 60 j m/s and v 80 i – 40 j m/s

find the time elapsed after the projectile is thrown up.

uy = 60, vy = –40

vy = uy – gt

–40 = 60 – gt

t = 10 s

Solution :

Instantaneous radius of Curvatureds approximates a circular arc.

t

2t

cosv v

cosv

gcosr

2 2

3

v cosr

gcos

vt

vtcos

mg mgcos

x

y

v

ds

Class Exercise - 4A ball is thrown with a velocity of m/s at an angle of 45° with the horizontal. It just clears two vertical poles of heights 90 cm each. Find the separation between the poles.

7 2

Solution :

P Q

A B

Y

XO

v

45°

Let t be the time after which the ball is at the top of the poles.

y(t) = 0.9 m

vy(0) = v sin45º 117 2 7 ms

2

Now, y(t) = vy(0)t + ayt212

0.9 = 7t – (9.8)t212

4.9t2 – 7t + 0.9 = 0

Solution

On solving, we get

1 21 9

t s or t s7 2

Hence the ball is at A after s and at B after s.

1

79

7

OP = vxt1 = 7 × = 1 m1

7

OQ = vxt2 = 7 × = 9 m97

PQ = 8 m

Class Exercise - 8

R

4

AB C D E

v = 40 m /s

A

x m/s

0 R

30°

R

2

3R

4

What should be the velocity of x such that the objects A, B would collide if both are thrown simultaneously?

Solution

For the particles to collide, vertical velocity should be the same. So, B should be thrown vertically up with a velocity of 20 m/s.

Projectile motion on an inclined plane

A particle is thrown horizontally from top of an inclined plane (inclination with horizontal) with speed . How far from the point of projection will it strike the plane

g

O u

Projectile motion on an inclined plane

Axis ox and oy.

Along x axis

initial x

x

2

v ucos

a gsin

1x OP ucos t gsin t ......(1)

2

initial y

y

2

v usin

a gcos

y O

1O ut sin gt cos ......(2)

2

Along y axis

y

x

g

u

Projectile motion on an inclined plane

T=0 (particle at O)

2

2usint (particle at P)

gcos

2u(i) OP tan sec

g

y

x

g

u

HOME EXERCISE

Illustrative Problem

What is the product of 2.6 0.5 cmand 2.8 0.5 cm ?

Illustrative Problem

(b) Divide the vector into two vectors along x’ and y’ which add up the give .

OP33333333333333

OP33333333333333

(a) Find component of displacement along x’ and y’ axis

OP33333333333333

(c) Are the answer of (a) and (b) are same ?

O x’

y’

P

30o

60o

OP = 50 Km

Illustrative Problem

A car is traveling at 10m/sec. driver sees a wall at a distance at 150m. he applies brakes at retardation of 1m/sec2. Find his distance from wall at t = 15 sec ?

Illustrative Problem

In Searle’s apparatus diameter of thewire was measured 0.05 only screw gauge of least count 0.001 cm. The length of wire was measured 110 cm by meter scale of least count 0.1 cm.an external load of 50 N was applied.the extension in length of wire was measured 0.125 cm by micrometer of least count 0.001 cm. find the maximum possible error in measurement of young’s modules.

F / A stressY where

/ strain

Y = young modules,A = area of cross section = R2 where R is radius.

Hints

Illustrative Problem

Figure shows x – t graph of a particle. Find the time t such that the averagevelocity of the particle during the period 0 to t is zero.

x in m

t in second

10 20

20

10

O

Illustrative Problem

A particle starts from a point A and travels along the solid curve shown in figure. findapproximately the position B of the particlesuch that the average velocity between thepositions A and B has the same direction asthe instantaneous velocity at B.

2 4 6 x

y

2m

4m

Thank you