(a) 1m s -2 (b) 2 m s -2 (c) zero (d) -1 m s -2

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11 th Physics - Unit 2 Kinematics Solutions for the Textbook Problems One Marks 1. Which one of the followin Cartesian coordinate system is not followed in physics? 5.

If the velocity is v = 2i + t 2 j 9k, then the manitude of acceleration at t = 0.5 s is (a) 1m s -2 (b) 2 m s -2 (c) zero (d) -1 m s -2 7.

4 m (c) 80.5 m (d) 79.2 m 8. A ball is projected vertically upwards with a velocity v. It comes back to round in time t. Which v-t raph shows the motion correctly? 2.

a) Mass b) Lenth c) Momentum d) Manitude of acceleration 4. Two objects of masses m1 and m2 fall from the heihts h1 and h2 respectively.

1 11 th Physics - Unit 2 Kinematics Solutions for the Textbook Problems One Marks 1. Which one of the followin Cartesian coordinate system is not followed in physics? 5. If a particle has neative velocity and neative acceleration, its speed (a) increases (c) remains same (b) decreases (d) zero 6. If the velocity is v = 2i + t 2 j 9k, then the manitude of acceleration at t = 0.5 s is (a) 1m s -2 (b) 2 m s -2 (c) zero (d) -1 m s If an object is dropped from the top of a buildin and it reaches the round at t = 4 s, then the heiht of the buildin is (inorin air resistance) ( = 9.8 ms -2 ) (a) 77.3 m (b) 78.4 m (c) 80.5 m (d) 79.2 m 8. A ball is projected vertically upwards with a velocity v. It comes back to round in time t. Which v-t raph shows the motion correctly? 2. Identify the unit vector in the followin. 3. Which one of the followin physical quantities cannot be represented by a scalar? a) Mass b) Lenth c) Momentum d) Manitude of acceleration 4. Two objects of masses m1 and m2 fall from the heihts h1 and h2 respectively. The ratio of the manitude of their momenta when they hit the round is 9. If one object is dropped vertically downward and another object is thrown horizontally from the same heiht, then the ratio of vertical distance covered by both objects at any instant t is (a) 1 (b) 2 (c) 4 (d) A ball is dropped from some heiht towards the round. Which one of the followin represents the correct motion of the ball?

R 60. Choose the correct relation from the followin Solution 11. If a particle executes uniform circular motion in 1.

(a) +y direction (b) +z direction x y =z usin (cross product rule) (c) -z direction (d) -x direction 2. Option D 12.

a) i j = 2 1 (b) The acceleration and speed are constant. (c) The velocity and acceleration are constant. (d) The speed and manitude of acceleration are constant. 13.

2 R 60. Choose the correct relation from the followin Solution 11. If a particle executes uniform circular motion in 1.Option D the xy plane in clock wise direction, then the anular velocity is in In Physics we prefer riht handed coordinate system. The condition to check it is as follows. (a) +y direction (b) +z direction x y =z usin (cross product rule) (c) -z direction (d) -x direction 2. Option D 12. If a particle executes uniform circular motion, choose the correct statement Manitude of unit vector = 1 (a) The velocity and speed are constant. a) i j = 2 1 (b) The acceleration and speed are constant. (c) The velocity and acceleration are constant. (d) The speed and manitude of acceleration are constant. 13. If an object is thrown vertically up with the initial speed u from the round, then the time taken by the object to return back to round is b) i 2 = c) k j 2 = d) i +j 2 = =1 3. Option C Momentum is a vector hence it cannot be represented by a scalar. 4. Option C Since they are fallin, initial velocity is zero. 14. Two objects are projected at anles 30 and 60 respectively with respect to the horizontal direction. The rane of two objects are denoted as R 30 and u1= u2=0 v 2 =u 2 +2as v1 2 =2h1

3 v2 2 =2h2 m 1 v 1 m 2 v 2 = m 1 2h 1 m 2 2h 2 5. Option A Neative acceleration increases the neative velocity whereas it decreases positive velocity. 6. Option A dv = a; a = d(2i +t2 j 9k ) = 2t j ; dt dt a = 2t; a t=0.5 = 1ms 2 7. Option B s = ut at2 ; since the object is dropped u = 0, a = 9.8ms 2. s = 1 (9.8)(4) = 78.4m Option C Clockwise direction is conventionally taken to be neative and perpendicular to the plane in which circular motion takes place (Check Cross Product Rule). 12. Option D A circular motion always posses centripetal acceleration whose direction chanes continuously with time thouh it s radial, therefore only possible choice is D. 13. Option C (Time of Fliht) T f = 14. Option A (Rane) R f = u2 sin 2θ 2u sin θ ; Here θ = 90 ; sin 60 = sin Option C 15. Option B Graph C alone matches our scenario. s = ut at2 ; u=0 a = 2s t 2 = 2(50) = 25ms 2 4 Exercise Problems 1. The position vectors particle has lenth 1m and makes 30 with the x-axis. What are the lenths of the x and y components of the position vector? 9. Option A Since horizontal velocity has no effect on vertical direction and distance covered by both the objects are same (action of ravity only). 10. Option A Due to ravity the velocity increases with time hence distance travelled by ball in successive equal time interval increases.

x = r cos θ = r cos 30 = (1) 3 2 y = r sin θ = (1)sin30 = 1 2 2. A particle has its position moved from r 1 = 3i + 4j to r 2 = i + 2j.

4 x = r cos θ = r cos 30 = (1) 3 2 y = r sin θ = (1)sin30 = A particle has its position moved from r 1 = 3i + 4j to r 2 = i + 2j. Calculate the displacement vector r and draw the r 1, r 2 and r in a two dimensional Cartesian coordinate system. 5. What are the resultants of the vector product of two iven vectors iven by A = 4i 2j + k and B = 5i + 3j 4k? i j k A B = = i (8 3) j ( 16 5) + k ( ) = 5i + 20j + 22k 6. An object is projected at an anle such that the horizontal rane is 4 times of the maximum heiht. What is the anle of projection of the object? Given R = 4h max 2u 2 sin θ cos θ = 4 u2 sin 2 θ 2 r = r 2 r 1 = i + 2j ( 3i + 4j ) 1 = tan θ = i + 2j 3i 4j = 2i 2j. tan 1 1 = θ 3. Calculate the averae velocity of the particle whose position vector chanes from r 1 = 5i + θ = 45 6j to r 2 = 2i + 3j in a time 5 second. 7. An object at an anle such that the horizontal rane is 4 times of the Unit 2 Kinematics 101 maximum heiht. What is the anle of projection of the object? v = r t = r 2 r 1 = t = 3 (i + j ) 5 2i + 3j 5i 6j 5 4. Convert the vector r = 3i + 2j into a unit vector. r = r r = 3i + 2j 3i + 2j =

In a v-t raph, slope ives acceleration. Usin the slope of each raph followin conclusions have been made. 9. A particle is projected at an anle of θ with respect to the horizontal direction.

5 In a v-t raph, slope ives acceleration. Usin the slope of each raph followin conclusions have been made. 9. A particle is projected at an anle of θ with respect to the horizontal direction. Match the followin for the above motion. a) Slope is constant (a straiht line ) hence acceleration is constant. b) Slope is zero, hence acceleration is zero. c) Slope is constant since the line is more steeper than a, acceleration is constant but a reater value than a. d) Slope increases with time hence acceleration is variable. 8. The followin velocity time raph represents a particle movin in the positive x direction. Analyse its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s. (a) vx decreases and increases (b) vy remains constant (c) Acceleration varies (d) Position vector remains downward vx = remains constant Explanation: For a particle bein projected the acceleration is alon vertical due to ravity and hence horizontal velocity remains constant vy = decreases and increases Explanation: Due to neative acceleration the positive velocity decreases until it becomes zero at maximum heiht and from then on it increases in neative direction. a= remains downward Explanation: Acceleration due to ravity. r = varies Explanation: It is based on the choice of oriin. 10. A water fountain on the round sprinkles water all around it. If the speed of the water comin out of the fountain is v. Calculate the total area around the fountain that ets wet. In a v-t raph area under curve ives the displacement. Distance = 1.75m and Displacement = m

In water fountain, the water is projected with initial velocity v and the maximum rane occurs when θ = 45 and the total area around the fountain that ets wet is a circle whose radius is Rmax.

6 In water fountain, the water is projected with initial velocity v and the maximum rane occurs when θ = 45 and the total area around the fountain that ets wet is a circle whose radius is Rmax. R = v2 sin 45 = v2 Area = πr 2 = πv The followin table ives the rane of a particle when thrown on different planets. All the particles are thrown at the same anle with the horizontal and with the same initial speed. Arrane the planets in ascendin order accordin to their acceleration due to ravity, ( value). Takin square both sides A² + B² + 2A.Bcosθ = B²/4 A² + 2ABcosθ + 3B²/4 = 0 Also, A and R is perpendicular with each other, A.(A + B) = 0 A.A + A.B = 0 A² + A.Bcosθ = 0 cosθ = - A/B A² + 2A.B(-A/B) + 3B²/4 = 0 A² - 2A² + 3B²/4 = 0 A = 3B/2 Now, cosθ = -A/B = - 3B/2B = - 3/2 cosθ = cos150 Ф = 150 Hence anle between A and B = Compare the components for the followin vector equations a) Tj mj = maj b) T + F = A + B c) T F = A B d) Tj + mj = maj R = u2 sin 2θ ; R 1 jupiter > earth > mars > mercury 12. The resultant of two vectors A and B is perpendicular to vector A and its manitude is equal to half of the manitude of vector B. Then the anle between A and B is Manitude of resultant of A and B is iven by, R = (A² + B² + 2ABcosθ) Accordin to the question, Manitude of resultant = half of Manitude of B (A² + B² + 2A.Bcosθ) = B/2 a) T m = ma b) T x + F x = A x + B x 14. Calculate the area of the trianle for which two of its sides are iven by the vectors A = 5i 3j, B = 4i + 6j. Area of a trianle whose sides are vector A and B 1 2 A B i j k A B = = i (0) j (0) + k ( ) = 42k

1 2 A B = 1 2 (42)2 = 21 15. If Earth completes one revolution in 24 hours, what is the anular displacement made by Earth in one hour. Express your answer in both radian and deree.

7 1 2 A B = 1 2 (42)2 = If Earth completes one revolution in 24 hours, what is the anular displacement made by Earth in one hour. Express your answer in both radian and deree. Earth completes one revolution in 24 hours. Hence anular displacement in 24 hours = 360 anular displacement in 1 hour =360/24=15. Hence, anular displacement made by earth in one hour = 15 The relation between radian and deree, 360 = 2π radian u = 20ms 1 ; θ = 30 R = u2 sin 2θ = 400 ( 3 2 ) 9.8 = 35.3m 1 = 2π/360 radian = π/180 which is lesser than the distance at which post is located hence the ball will reach it. 15 = 15π/180 = π/12 radian. 16. A object is thrown with initial speed 5 ms If an object is thrown horizontally with an initial with speed 10 m s -1 from the top of a buildin of heiht an anle of projection 30. What is the heiht and 100 m. what is the horizontal distance covered by rane reached by the particle? the particle? u = 5ms 1 ; θ = 30 h max = u2 sin 2 θ 2 = 25 (1 2 ) = 0.318m Since the object is horizontally projected its vertical velocity is zero. u x = 10ms 1 ; u y = 0 R = u2 sin 2θ = 25 ( 3 2 ) = 2.21m A foot-ball player hits the ball with speed 20 ms -1 with anle 30 with respect to horizontal direction as shown in the fiure. The oal post is at distance of 40 m from him. Find out whether ball reaches the oal post? We resolve the motion into horizontal and vertical. s y = u y t t2 2s y = t = = 4.5s s x = u x t = = 45m

8 20. Consider the x axis as representin east, the y axis as north and z axis as vertically upwards. Give the vector representin each of the followin points. a) 5 m north east and 2 m up b) 4 m south east and 3 m up c) 2 m north west and 4 m up

9 21. The Moon is orbitin the Earth approximately once in 27 days, what is the anle transversed by the Moon per day? Anle transversed by Moon in 27 days = 360 Anle tranversed in one day = = An object of mass m has anular acceleration α = 0.2rad s 2 What is the anular displacement covered by the object aft er 3 second? (Assume that the object started with anle zero with zero anular velocity). ω i = 0; t = 3s; θ = ω i t αt2 = 1 (0.2)9 = 0.9rad 2 Prepared by E. Arun Moses M.Sc. Tutor at J.F Educational Service. reachjamics@mail.com Please write to us for further development.

XI PHYSICS M. AFFAN KHAN LECTURER PHYSICS, AKHSS, K. https://promotephysics.wordpress.com

XI PHYSICS M. AFFAN KHAN LECTURER PHYSICS, AKHSS, K affan_414@live.com https://promotephysics.wordpress.com [MOTION IN TWO DIMENSIONS] CHAPTER NO. 4 In this chapter we are oin to discuss motion in projectile