CLS Aipmt 16 17 XI Phy Study Package 1 SET 1 Chapter 4

Chapter

4

Motion in a Plane Solutions SECTION - A School/Board Exam. Type Questions Very Short Answer Type Questions : 1.

Which of the following is a scalar quantity? Momentum, Acceleration, Work, Force.

Sol. Work is a scalar quantity, others are vectors. 2.

Name two vector quantities.

Sol. Displacement and velocity. 3.

Can three vectors of different magnitudes be combined to give a zero resultant?

Sol. If three vectors form the sides of a triangle when taken in the same order, the resultant vector is a null vector. 4. If | A  B |  | A  B |, what is the angle between A and B ?

Sol. Let the angle between A and B be . | A  B |  A2  B 2  2 AB cos  | A  B |  A2  B 2  2 AB cos   A2 + B2 + 2ABcos = A2 + B2 – 2ABcos  4ABcos = 0  = 90° 5.

What is the average value of acceleration of an object in uniform circular motion in one complete revolution?

v Sol. a  t

In one complete rotation v  v  v  0 . On reaching the same point, the object has same velocity.  6.

a 0

Electric current has both magnitude and direction, so is it a vector?

Sol. No, electric current does not obey vector laws of addition. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

58 7.

Motion in a Plane

Solution of Assignment (Set-1)

Can the sum of two vectors be a scalar?

Sol. No, sum of two vectors is always a vector. 8.

What is the magnitude of i  j ?

Sol. The vector is i  j  Its magnitude = =

12  12

2

What is the direction of A  B for parallel vectors A and B ? Sol. It is along either A or B , both have same direction.

9.

10. When can the sum of two vectors be minimum and maximum? Sol. Minimum, when two vectors are in opposite directions. Maximum, when two vectors are parallel i.e., in same direction. Short Answer Type Questions : 11. Two vectors having magnitudes A and resultant and A ? Sol. Let R be their resultant. Then by law of cosines | R | A2  ( 3 A)2 | R |  2A Using law of sines,  sin  

sin  

3 A are perpendicular to each other. What is the angle between their

R |R| 3A  sin90 sin 



3A

A

3A  sin90 2A

3 2

 = 60° 12. Find the angle of the vector A  3i  6 j  kˆ with y-axis.

Sol. Here | A |  33  62  ( 1)2 =

9  36  1

= 46 Let A make angle  with y-axis, then | A | cos   6  cos  

6

 0.88465 46  = cos–1 (0.88465)  = 27.79° Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


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13. A boy sitting in a train moving with constant velocity, throws a ball vertically upwards. How does the ball appear to move to an observer, (i) inside the train, (ii) outside the train. Sol. (i) To an observer sitting inside the train, the ball will appear to move straight vertically upwards and then downwards. (ii) To an observer sitting outside the train, the ball will appear to move along the parabolic path. 14. Rain is falling vertically with speed 20 m/s. A man runs with speed of 10 m/s towards east. In which direction should he hold his umbrella? Up Sol. Velocity of rain with respect to man

v rm  v rain  v man 10 20 1 = 2

 tan  

–vman

West



vrain

10 m/s East vman

20 m/s

Down

Or = 26.56° inclined to vertical towards east direction. 15. Are the two vectors (2 kg) (4 m/s, towards east) and 2(4 m/s, towards east) same?

Sol. No, (2 kg) (4 m s–1, towards east) shows the multiplication of a scalar quantity i.e., mass of 2 kg with a velocity vector. So it is a momentum vector of magnitude 8 kg m/s directed towards east. 2 (4 m/s, towards east) is the velocity vector of magnitude 8 m/s. Hence, the two are not same. 16. If A  B  C and | C |  | A | and | B | . Does that mean | C |  | A |  | B | ? Sol. No, In above case, A, B and C represent three sides of a triangle. And no side is ever greater than the sum of the other two. Hence, | C |  | A |  | B | 17. Show graphically that subtraction of two vectors is not commutative. Sol. The two figures given show that A  B and B  A are two different vectors.

B

–A

A A–B

B

B–A

A –B

Hence, subtraction of vectors is not commutative. 18. A projectile is fired with kinetic energy 4 kJ. If its range is maximum, what is its K.E. at the highest point of its path? Sol. Since the range is maximum, the angle of projection is 45°. If velocity of projection = v  4 kJ 

1 mv 2 2

Velocity at the highest point = v cos45° =  K.E. at highest point =

1 ⎛ v ⎞ m⎜ ⎟ 2 ⎝ 2⎠

v 2

2

Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Motion in a Plane




1⎛ 1 ⎞ mv 2 ⎟ 2 ⎜⎝ 2 ⎠

4 kJ  2 kJ 2 19. For a given velocity in projectile motion, name the quantities related to a projectile which have maximum values when the maximum height attained by the projectile is the largest. 

Sol. (i) Vertical component of initial velocity (ii) Angle of projection (iii) Time of flight 20. Give a few examples of motion in two dimensions. Sol. (i) Projectile motion (ii) Uniform circular motion (iii) Non-uniform circular motion : The object moving in a circle has different speeds at different points. For example, A stone tied with a thread whirled in a vertical plane. 21. Find the angle of projection at which the horizontal range and maximum height of a projectile are equal. Sol.  Hmax = R 

v 02 sin2 0 v 02 sin 20  2g g

=

sin 0  2cos 0 2

 tan0 = 4 or 0 = 75.96° 22. Why is centripetal acceleration called so? Is it a constant vector? Sol. Centripetal acceleration always acts towards the centre of the circular path at every point. Hence, it is called centripetal acceleration which in Greek means “center seeking”. Since its direction changes continuously, it is not a constant vector. 23. What do you mean by resolution of a vector and components of a vector? Sol. Splitting a vector into two or more component vectors is called resolution of the vector. A vector can be resolved into infinite number of components. However a vector can have only two rectangular components in a plane. Components of vector : If vectors A, B and C, when added using vectors laws, result into vector D , then they are said to be component vectors of D along directions A, B and C . 24. Find components of vector addition of 2i  4 j  k and 3i  2 j  2k . Sol. Resultant

R  (2i  4 j  k )  (3i  2 j  2k )

R  (5i  2 j  k )

Hence, rectangular components of R are Rx = 5 along x-axis Ry = 2 along y-axis Rz = 1 along z-axis Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


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25. If the position of particle at time t is given by r  2t 2 i  6t j  8k. r has the unit m. Find the component of velocity

along z-axis at time t = 4 s. Sol. r  2t 2 i  6t j  8k

dr v  4ti  6 j  0k dt = 4ti  6 j

∵ vz = 0, here. The particle does not have any velocity along z-axis. 26. An object is projected at an angle 30°. If its horizontal velocity is 50 km/h, what is its vertical velocity? Sol. 0 = 30° v0cos0 = 50 km/h v0sin0 = ? v 0 sin 0 

50  sin30 cos30

= 50 

1

km/h = 28.86 km/h 3 27. What is the relative velocity of a man swimming downstream with speed 12 km/h (in still water) with respect to a child running towards the river with speed 4 km/h in direction perpendicular to water flow. Speed of water flow = 4 km/h. Sol. Velocity of man in still water = 12 km/h = v m

Velocity of water flow v w  4 km/h  Net velocity of man w.r.t. ground = v mn  16 km/h i Velocity of child = vC  4 km/h j  Velocity of man w.r.t. child v ( mn )c  16i  4 j

| v ( mn )c |  (16)2  (4)2 = 4 17 km/h 1 ⎛ 1 ⎞

  tan ⎜ ⎟ ⎝ 4⎠

vw vmn vC

y O

x

= – 14.04° with x-axis (i.e., direction of flow of river). 28. What do you mean by angular speed in uniform circular motion? How is it related to time period and centripetal acceleration? ) : The time rate of change of the angular displacement of an object having uniform circular motion Sol. Angular speed ( is called angular speed (). 

2 or aC = 2r T


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Motion in a Plane


29. A cyclist going round a circular path with constant speed 10 km/h completes 42 revolutions in 30 minutes. Find its centripetal acceleration. Sol. Frequency   

7 1 42 s = 30  60 300

Time period T 

Also T 

 R

...(i)

1 300  s  7

2R v

vT 2

Given that v = 10 km h–1

 R

5 300  18 7 m  625 m 22 33 2 7

10 

...(ii)

aC = 4 22R = 4× =

22 22 7 7 625     [using equation (i) and (ii)] 7 7 300 300 33

11 m/s2 27

= 0.41 m s–2 30. Find time of flight of an object projected at angle 30° with speed 60 m/s. [take g = 10 m/s2 ] Sol. T 

2  60  1 2v 0 sin 0 = 10  2 g

=6s

Long Answer Type Questions : 31. State and prove the law of cosines. Sol. Law of cosines : The resultant R of two vectors P and Q inclined at an angle  is given by

R  P 2  Q 2  2PQ cos  .

Let the two vectors P and Q are inclined at an angle  with each other. Then, the resultant of their vector addition can be obtained by using parallelogram method as shown below. Here, AB  P  Q From B, draw a line perpendicular to AC which meets it at E when extended. Now in ABE AB2 = AE2 + BE2

B [Pythagoras theorem]

 AB2 = (AC + CE)2 + BE2 From BCE, we can have

Q

...(i)

A



P+

 P

Q

 C

E



CE = CBcos = Qcos

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63

[CB = Q, opposite sides of a parallelogram]

Also BE = CBsin = Qsin Substituting these values in equation (i) above, AB2 = (P + Q cos)2 + Q2sin2 AB2 = P2 + Q2 + 2PQcos 

AB  P 2  Q2  2PQ cos 

Law of cosines

32. Define a unit vector. Explain how can we express a vector using unit vectors along coordinate axes? Find unit vector along the vector 6i  8 j  10k.

Sol. A vector having unit magnitude is called a unit vector. The unit vector along the direction of vector A is denoted by A A  is given by A. A . A It does not have any unit. It is convenient to express a general vector in terms of its rectangular components using unit vectors. If the components of a vector P along the respective axes are Px, Py, Pz. Then

P  Px i  Py j  Pz k i , j and k are unit vectors along x, y and z-axes. Let A  6i  8 j  10k

= | A |  62  82  102 =

200

= 10 2

A 6i 8 j 1k    Unit vector A   | A | 10 2 10 2 2 3i 4 j 1k   = 5 2 5 2 2 33. Explain how can we resolve a vector into two components lying in its plane? Sol. The process of splitting up of a vector into two or more vectors is known as the resolution of a vector. Let us consider the components in a plane. Suppose we want to resolve a vector P along any two directions, ˆ , which lie in a plane containing P . Here lˆ and m ˆ are the unit vectors along two directions. say lˆ and m ^ m ^

I

P Figure (i)

Now if we draw a straight line parallel to l from the tail of P and another straight line parallel to mˆ from the head of P , and mark their intersection point as N (say), then we obtain an arrangement as shown in figure Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Motion in a Plane


(ii) below. ON is a vector parallel to l , and NS is a vector parallel to mˆ .

S P O N Figure (ii)

Then ON  l ˆ And NS  m

Whereand are the real numbers denoting the magnitude of ON and NS respectively. By applying triangle method of vector addition in figure (ii). We get P  ON  NS

ˆ  P  lˆ  m

. Or we say and are the components of vector P along the directions l and m 34. (i) Derive expression for the horizontal range of a projectile,

2

times the actual range for a given velocity of projection. What 3 is the angle of projection for the actual range?

(ii) The maximum range of a projectile is

Sol. (i) The maximum horizontal distance travelled by the projectile during its flight is called the horizontal range of the projectile. This is the straight distance OP as shown in figure. It is denoted by R. R can be calculated by using equation x = (v0cos0)t When x = R, t  Time of flight, Tf 



R  (v 0 cos 0 )

R

v 02 sin 20 g

2v 0 sin 0 g

(2v 0 sin 0 ) 2v 02 (sin 0 )(cos 0 )  g g

[∵ 2 sin0 cos0 = sin(20)]

R is maximum when 20 = 90° i.e., Rmax = (ii) Rmax 

v 02 2  R g 3

y v0 O

0 R

P x

v 02 g

[R = actual Range]

v 02 2 ⎛ v 02 sin20 ⎞  ⎜ ⎟⎟ g g 3 ⎜⎝ ⎠

sin20 

3 2

 20 = 60°  0 = 30° Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


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35. Can the resultant of three non parallel, coplanar forces of magnitudes 4 N, 8 N and 3 N acting on a particle be zero? Explain. Sol. A vector having zero magnitude and having any arbitrary direction is called a null vector. It is denoted by 0 .

Examples of null vector : (i) Zero displacement (ii) Zero change in velocity (iii) Resultant of two equal and opposite forces acting at a point. Properties of null vector 0. A0 A A0 A A .0  0 A .0  0 Vector sum of three non-parallel and coplanar vectors will be zero only when the given vectors are represented by three sides of a triangle. From geometry, the sum of any two sides of a triangle must be greater, than the third side. Here in given problem (4 + 3) N < 8 N. The forces can not be represented by the side of a triangle. Hence their resultant can not be zero. 36. (i) Show that a projectile follows a parabolic path. (ii) Find the horizontal distances travelled by a projectile projected at an angle 30° with horizontal with speed 15 m/s, at the time it is at height 2.5 m above the point of projection. [take g = 10 ms–2] Sol. (i) Consider a stone projected with velocity v0 at angle  with x-axis. v0x = v0 cos

[Horizontal speed at the time of release]

v0y = v0 sin

[Vertical speed at the time of release]

Along horizontal :  v x  v 0 x  v 0 cos i

y

If we take the time at which the stone is projected as t = 0, then its horizontal displacement at any time t is xi  (v 0 cos  )tiˆ

or x = v0cost

v0y

v0

 O v0x

a = – gj

x

...(i)

Along vertical : The stone moves under a constant acceleration a  g j . The velocity v y of the stone at time t is given by

v y j  v 0 y j  ( g )t j

 vy = v0sin – gt

Its vertical displacement y in this time is given by y  v 0 sin t 

1 2 gt 2

Substituting t 

x from equation (i) v 0 cos 


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Motion in a Plane


⎞ ⎛ ⎞ 1 ⎛ x x2 y  v 0 sin  ⎜ ⎟⎟ ⎟  g ⎜⎜ 2 2 ⎝ v 0 cos  ⎠ 2 ⎝ v 0 cos  ⎠ 

y   tan   x 

1 g x2 2 (v 02 cos2 )

Equation of trajectory of the projectile.

(ii) v0 = 15 m/s  = 30° y  (v 0 sin )t 

1 2 gt  2.5 m (vertical height) 2

2.5  (15 sin 30 )t 

2.5 

1  10t 2 2

15 t  5t 2 2

 10t2 – 15t + 5 = 0  t = 0.5 s, 1 s Horizontal distance travelled x = v0cos0t = 15 cos 30  0.5 [Taking t = 0.5 s] = 15 

2.5 m

3  0.5  6.5 m 2

2.5 m 12.99 m

Similarly for t = 1 s, x = 12.99 m

37. Prove the commutative and associative properties of vector addition. How do we specify the position of an object using vectors?

Sol. (i) Commutative Property : To obtain A + B , we shift B parallel to itself so that its tail coincides with the head of A . The line segment joining the tail of A to the head of A represents A + B and its direction is as shown in the figure (ii).

B B

A+B B

A (i)

A (ii)

If instead of B , we shift A parallel to itself such that the tail of A coincides with the head of B , the vector obtained by joining the tail of B to the head of B gives B + A . A B A

B+A (iii)

If we compare A + B and B + A from figures (ii) and (iii), we find the two vectors are equal i.e., A + B  B + A . Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


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(ii) Associative Property : The figure shows that A  B  C  A  (B  C )  ( A  B )  C

A+ B+ C

A A

B

+ +

B

B

C

C We can describe the position of an object using position vector. The position vector of a point P is the vector joining origin O to the point P. Thus it is the line OP having arrow at P. 38. The position of an object is described by the vector r  t 2 i  2t j  tk at any time t. Find its position, velocity and acceleration at time t = 6 s. Sol. r  t 2 i  2t j  tk

dr v  2ti  2 j  k dt dv a  2i  0 j  0k dt at t = 6 s, (i) Position :

r  62 i  12 j  6k = 36i  12 j  6k

(ii) Velocity :

v  2  6i  2 j  1k = 12i  2 j  k

| v | 122  22  ( 1)2 = (iii) Acceleration :

149 m s–1

a  2i

A constant acceleration of 2 m s–2 along positive x-axis. 39. If A  2iˆ  jˆ and B  iˆ  2 jˆ , find the scalar magnitude and directions of (i) A (ii) B and (iii) A  B Sol. (i)

A  2iˆ  jˆ Magnitude of A  (2)2  ( 1)2  5

Direction of A with x-axis. ⎛ 1⎞  = tan–1 ⎜ ⎟ = –26.56° ⎝ 2⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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(ii) B  iˆ  2 jˆ Magnitude of B  (1)2  ( 2)2  5 Direction of B with x-axis

⎛ 2 ⎞  = tan–1 ⎜ ⎟ = –63.43° ⎝ 1⎠

(iii) A  B  (2iˆ  jˆ)  (iˆ  2 jˆ)  (3iˆ  3 ˆj )  Magnitude of A  B  32  ( 3)2  3 2

Direction of A  B with x-axis. ⎛ 3 ⎞  = tan–1 ⎜ ⎟ = –45° ⎝ 3⎠ 40. Show that a two-dimensional uniformly accelerated motion is a combination of two one-dimensional motions along perpendicular directions. Sol. A body is said to be moving with uniform acceleration if its velocity vector suffers the same change in equal interval of time, however small. For this case the average acceleration of the object is same as its instantaneous acceleration over a given time interval. The motion in a plane with uniform acceleration can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. This can be shown as follows. Let a constant acceleration a act on an object moving in a plane. This acceleration changes its velocity from v 0 at time t = 0 to v at time t = t. Then v  v0 a t 0  v  v 0  at ...(i) As we have studied in the last chapter, that for an object having constant acceleration, average velocity is given as v0  v v  ...(ii) 2 r  r0 From the definition of average velocity during the time interval t = t – 0, it can be expressed as v  . t 0 Where r and r 0 are the position vectors of the particle at time t = t and t = 0 respectively.  vt  r  r0 or r  r0  vt ⎛ v  v ⎞ r  r0  ⎜ 0 t [From equation (ii)] ⎜ 2 ⎟⎟ ⎝ ⎠ ⎛ v  v  at ⎞ 0 0 ⎟⎟ t [From equation (i)]  r  r0  ⎜⎜ 2 ⎝ ⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


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1 r  r0  v 0t  at 2 2

Writing in component form 

1 xi  y j  x0 i  y 0 j  (v 0 x i  v 0y j )t  (ax i  ay j )t 2 2

Rearranging 

1 1 ⎛ ⎞ ⎛ ⎞ xi  y j  ⎜ x0  v 0 x t  ax t 2 ⎟ i  ⎜ y 0  v 0 y t  ay t 2 ⎟ j 2 2 ⎝ ⎠ ⎝ ⎠

Comparing two sides 1 ax t 2 2 1 y  y 0  v 0 y t  ay t 2 2 x  x0  v 0 x t 

along x-axis ⎫ ⎬ ...(iii) along y-axis ⎭

Thus we see that the motions in x and y-directions can be treated independently from each other. This result simplifies our study of motion in a plane. A similar result can be obtained for the three-dimensional motion of an object. So for this case, we get a set of three equations as follows.

x  x0  v 0 x t 

1 ax t 2 2

y  y0  v0y t 

1 ay t 2 2

z  z0  v 0 z t 

1 2 az t 2

41. Define the following, (i) Relative velocity (ii) Average velocity (iii) Average acceleration (iv) Centripetal acceleration Sol. (i) Relative velocity : Let two objects A and B are moving in a plane with velocities v A and v B measured with respect to ground. The relative velocity of A with respect to B is given by ...(i) v AB  v A – v B

Similarly relative velocity of B with respect to A is given by ...(ii) v BA  v B – v A Comparing (i) and (ii) v AB  v BA And | v AB |  | v BA | Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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(ii) Average velocity : Average velocity v of an object is the ratio of its displacement and the corresponding time interval.



r x i  y j v  t t t

r , the direction of average velocity is the same as that of r . Since v  t The instantaneous velocity or simply the velocity is the limiting value of the average velocity when the time interval approaches zero. (iii) Average Acceleration : The average acceleration a is the ratio of the change in velocity and the corresponding time interval. If the velocity of an object changes from v to v1 in time interval t, then the acceleration of the object is given by the relation v   v a t v  a [where v  v   v ] t

From the above relation, we can see that a is along the direction of v . The direction of v is different from that of v  and v as long as the object moves along a curve and not along a straight line. Or we say that for the motion along a curve, the direction of average acceleration is different from that of the velocity of the object. They may have any angle between 0° and 180° between them. (iv) Centripetal Acceleration : Consider a particle moving on a circular path of radius r and centre O, with a uniform speed v, as shown in the figure below. Let the particle be at point P at time t, and at Q at time t +t. Let v1 and v 2 be the velocity vectors at P and Q directed along the tangents at P and Q respectively. To find the change in velocity, v 2  v1  v

v2 This is the change in velocity during this time interval t.

r2 O

Q r



r1

v1 P

By definition, the average acceleration is given by v a i.e., a is along v. t

As t  0, the average acceleration becomes the instantaneous acceleration and  also approaches zero. Thus v and hence, a is perpendicular to velocity vector v 1 . But since v 1 is directed along tangent at point P, so acceleration ac acts along the radius towards the centre of the circle. That is why this acceleration is called centripetal acceleration which means ‘centre seeking’. Numerically its value is

v2 . r



Motion in a Plane

71

42. (i) Using graphical method, find the angle between (i  j ) and (i  j ) . (ii) Express the unit vector along above mentioned vectors.

y i+j 45° 45°

Sol. (i)

x

i –j Angle that iˆ  jˆ makes with x-axis

tan 1 

1 1

1 = 45° Similarly, the angle that iˆ  jˆ make with x-axis. tan 2  

1 1

2 = –45°  Required angle = 90° (ii) | i  j |  12  12 =

2

 

1 ij Therefore unit vector along i  j , is = 2 =

1 1 i j 2 2

Similarly | i  j |  2

 

1 1 1 i j ij =  Unit vector along i  j , is = 2 2 2 43. The ceiling of a long hall is 30 m high. What is the maximum horizontal distance that a ball thrown with a speed 45 ms–1 can go without hitting the ceiling of the wall? [take g = 10 m/s2] Sol. v0 = 4 5 ms–1 Let 0 = angle for projection Taking maximum height reached Hmax 

v 02 sin2 0 2g

Taking maximum height = 30 m We have

v 02 sin2 0  30 2g


72

Motion in a Plane 2  sin 0 


30  2  10 v 02

 sin2 0 

30  2  10 45  45

2  sin 0 

600 8  45  45 27

2 2 3 3

 sin 0 

...(i)

cos 0  1  sin2 0 =

1

 cos 0 

8  27

15 15 5   27 3 3 3

5 3

..(ii)

1 ⎛ 2 2 ⎞ Horizontal range R for 0  sin ⎜⎜ ⎟⎟ ⎝3 3 ⎠

R

v 02 sin 20 g

=

v 02 2 sin 0 cos 0 g

=

v 02 2 2 5  2  [Using equation (i) and (ii)] g 3 3 3



45  45 2  2 2   5 10 9 3

= 164.3 m 44. A body of mass 10 kg revolves in a circle of diameter 0.8 m completing 420 revolutions in a minute. Calculate its (i) Angular speed (ii) Linear speed (iii) Time period and (iv) Centripetal acceleration Sol. Given, Diameter = 0.8 m  Radius R = 0.4 m

Frequency ( ) =

420 Number of revolutions = rev/s = 7 s–1 60 Time taken

(i) Angular speed  = 2

= 2

22  7 = 44 rad/sec 7



T = time period 

v  2

73

2R T

(ii) Linear speed v 



Motion in a Plane

1 1  s  7

22  0.4 44  0.4 = 0.3591 m s–1 = 49 77

(iii) Time period T 

1 = 0.142 s 7

(iv) Centripetal acceleration aC = 2R = (44)2 × 0.4 = 774.4 m/s2 45. A bird is flying with velocity 5iˆ  6 ˆj w.r.t. wind. Wind blows along y-axis with velocity v. If bird is initially at A, and after sometime reaches B as shown. Find v, and also find the velocity of bird with respect to ground.

y

B 30°

A

x

Sol. Given that velocity of bird with respect to wind is v bw  5i  6 j Velocity of wind w.r.t. ground v wG  v j Now v bw  v bG  v wG

...(i)

5i  6 j  v bG  vwG  v bG  5i  6 j  v j v bG  5i  (6  v ) j

v bG is shown by the vector AB in figure. Since velocity v bG makes angle 30° with y-axis, it is inclined at 60° with x-axis. Hence tan60 

6v 5

3

6v 5



 v = 2.66 m s–1  Velocity of bird with respect to ground

v bG  5iˆ  6 ˆj  2.66 jˆ  5iˆ  8.66 jˆ


74

Motion in a Plane


SECTION -B Model Test Paper 1.

Name two vector quantities.

Sol. Force, acceleration 2.

Pick out the scalar quantity among the following : relative velocity, viscous drag, current, work, momentum.

Sol. Current, work 3.

If an object having uniform circular motion undergoes an angular displacement , what is the angle between its initial and final velocity vectors?

Sol.  4.

Give an example of null vector.

Sol. Zero displacement. 5.

Add the vectors 2iˆ  3 jˆ, iˆ  jˆ and  3 iˆ  2 jˆ .

Sol. Let a  2i  3 j b  1i  1j c  3i  2 j a  b  c  (2  1  3)i  (3  1  2) j = 0i  0 j = Null vector 6.

If a projectile is projected at 55°, at what other angle can it be projected to obtain the same range if projected with same speed?

Sol. 35° 7.

When can A  B be equal to A  B ?

Sol. When B  O

A  B  A and A  B  A 

AB  AB

When B is a null vector. 8.

What is the time period of an object completing 40 revolutions in a minute?

Sol.  

40 –1 2 –1 s  s 60 3

Time period T 

1 3   1.5 sec  2



9.

Motion in a Plane

75

Keeping the angle of projection same, how does the horizontal range of a projectile vary, when its initial velocity is doubled?

Sol. Horizontal distance R 

v 02 sin 20 g

...(i)

Let v 0  2v 0 

R 

=

(v 0 )2 sin 20 g

4v 02 sin20 g

R = 4R [From equation (i)] 10. Give two methods to specify a vector in a plane. Sol. A vector P can be specified by two methods.

(i) By its magnitude | P | and its orientation  with x-axis in x-y plane. (ii) By its rectangular components along the corresponding co-ordinate axes i.e., Px and Py.

 P  Px i  Py j 11. Find a unit vector parallel to the vector 3i  7 j  4k. Sol. Let A  3i  7 j  4k

| A |  3 2  72  42



 A

=

9  49  16

=

74

A A  | A|

1 74

(3i  7 j  4k )

12. Find the resultant of two forces of magnitude 12 N and 7 N acting simultaneously on a body at angle 60° with each other.

Sol. P  12 N Q 7N Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

76

Motion in a Plane


 = 60° R  P Q

Q

R  P 2  Q 2  2P cos 

60° P

1 R  12  7  2  12  7  2 2

2

=

144  49  84

=

277

= 16.64 N Let R makes with P Then

Q R  sin  sin 60

 sin  

=

[Law of sines]

Q sin 60 R

7 3  16.64 2

 sin = 0.364°   = 21.36° 13. Show that the horizontal range of a projectile is maximum when projected at angle 45°, for given v0. Sol. R 

v 02 sin 20 g

For given value of v0, R is maximum when sin20 = 1 i.e., 20 = 90° or 0 = 45°

14. The angle between velocities vA and v B is 60°. Find vA B if | vA |  16 and | v B |  4 Sol. Given

| v A |  16 | vB |  4

Choose x-y plane such that v B lies along x-axis. (See figure) Then v B  4i v A  (16cos 60)i  (16 sin 60) j

vA vAB

–vB O

60°

vB

x

= 8i  8 3 j By definition, relative velocity Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


Motion in a Plane

77

v AB  v A  v B

= (8i  8 3 j )  4i v AB  4i  8 3 j

Hence, | v AB |  (4)2  (8 3)2 =

16  64  3

= 14.42 m/s Let  be the angle of v AB with direction of v B i.e., direction of x-axis. 1 ⎛ 8 3 ⎞ ⎟⎟ Then   tan ⎜⎜ ⎝ 4 ⎠

= tan1(2 3) = 73.89° Hence, v AB is 28 m/s at an angle 73.89° with v B .

15. How would a projectile motion be affected if air resistance becomes too large to be neglected? Sol. If air resistance becomes too large, the path of the projectile will be different from its idealized trajectory. Maximum height attained and the horizontal range will be lesser than that attained in vacuum. 16. Find the magnitude of change in momentum of an oblique projectile during its whole journey if velocity and angle of projection are v0 and 0 respectively. Sol. Given that Velocity of projection = v0 Angle of projection = 0 In vector form, velocity at the time of projection v 0  v 0 cos 0 i  v 0 sin 0 j Velocity at the time of landing v   v 0 cos i  v 0 sin  j

[Projectile hits back with the same velocity, in downward direction]

 Change in momentum p  m(v   v 0 )

= m ⎡⎣(v 0 cos 0 i  v 0 sin 0 j )  (v 0 cos 0 i  v 0 sin 0 j )⎤⎦ p  2mv 0 sin 0 j So change in momentum is 2mv0sin0 in downward direction. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

78

Motion in a Plane


17. Find the magnitude of change in velocity for the uniform circular motion shown below.

4 m/s

O

B 60°

4 m/s A

Sol. Velocities at points A and B are v A  4 m/s normal to OA v B  4 m/s normal to OB

4 m/s

Since angle between OA and OB is 60°. Hence, angle between v A and v B is also 60°.  Change in velocity v  v B  v A

v

B

O

60°

4 m/s A

vB

° A 60 v

| v |  v A2  v B2  2v A .v B cos 60

=

16  16  2  16 

1 2

 v  4 m/s Along the perpendicular bisector of the angle AOB.

18. Define the following terms (i) Uniform circular motion (ii) Angular displacement (iii) Frequency of rotation Sol. (i) Uniform circular motion : An object is said to have uniform circular motion if it travels in a circular path with constant speed. (ii) Angular displacement : Angle that an object in circular motion subtends at the centre of its circular path, while going from one point to another, is called its angular displacement, in a given time. (iii) Frequency of rotation : Number of rotations completed in unit time is called the frequency of rotation. 19. Give one example of each of the following (i) A vector without unit (ii) A vector not having any specific direction (iii) A vector which is not constant Sol. (i) Unit vector does not have any unit or dimension. (ii) A null vector does not have a specific direction as its magnitude is zero. (iii) Centripetal acceleration is not a constant vector. Its direction changes continuously. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456


Motion in a Plane

79

20. Show that parallelogram method and triangle method of vector addition are equivalent.

Sol. Let A and B be two vectors. Their addition by triangle method and parallelogram method is shown below graphically.

A

+

B

B

B

A

+

B

A Parallelogram method

A Triangle method It is clear that the two methods bear the same result.

21. Find the velocity of an object after 3 s if its position at t = 0 is given as r  3ti  2t 2 j Sol. Given r  3ti  2t 2 j dr v   3i  4t j dt At t = 3, v  3i  12 j

| v |  (3)2  (12)2 =

153

Angle made with x-axis  = tan–1 (4) 22. State and prove the law of sines. Sol. Let A and B be two vectors, inclined at angle  with each other. Let their resultant R  A  B , make angle with A and angle  with B as shown in the figure. Draw OM  LN. Draw NP  LM which meets LM at point P when extended. In PNM sin  

PN PN  MN | B |

...(i)

PN PN In PNL, sin   LN  |R|

N

...(ii)



From (i) and (ii)

| B | sin  | R | sin  

|B| |R|  sin  sin 

In MOL, sin   In MON,

...(iii) OM OM  LM | A |

sin  

O

R

OM OM  MN | B |

L

 A

B  M

P

...(iv)

...(v)

From (iv) and (v), | A | sin  | B | sin  Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

80

Motion in a Plane




|A| |B|  sin  sin 

...(vi)

From, equation (iii) and (vi) we have, |A| |B| |R|    Law of sines sin  sin  sin  23. What do you mean by relative velocity? A boat goes in a river with speed 15 km/h (in still water) while a man crosses it with speed 8 km/h (in still water). The speed of water flow is 4 km/h and the boat goes downstream. How fast and in which direction does the man swim according to a man in boat? Sol. Relative velocity : The velocity of an object A as observed by another object B (moving or at rest) is called relative velocity of A with respect to B. v AB  v A  v B Where v A and v B are respective velocities with respect to a common reference point.

Let the direction of river flow be as shown in the figure. Given that Speed of boat in still water, vb = 15 km/h Speed of water flow vw = 4 km/h

Down stream flow y vw

As the boat goes downstream, net velocity of boat v b(net)  v b  vw

vm vm(net) x

ne

[Parallel vectors]

v mb(

t)

= 19i km/h

Velocity of man in still water = v m  8 j km/h [see figure]

Net speed of man in river, v m(net)  v m  v w

–vb(net)

vb(net)

River

= 8 j  4i Required velocity i.e., relative velocity of man w.r.t. boat v mb(net)  v m(net)  v b(net) = (8 j  4i )  19i = 8 j  15i  15i  8 j | v mb(net) |  ( 15)2  82

=

225  64

=

289

= 17 km/h

⎛ 8 ⎞ Direction   tan1 ⎜ ⎟ ⎝ 15 ⎠ = – 28.07° The man appears to swim at speed 17 km/h making angle 28.07° with the upstream.


CLS Aipmt 16 17 XI Phy Study Package 1 SET 1 Chapter 4

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