PHY 110 Handout I. Outline. Introduction

Transcription

1 Introduction PHY 110 Handout I Phsics is inherentl a science of measurement. It s a science dedicated to the stud of all natural phenomena. Phsics is a science whose objective is to stud the components of matter and their mutual interactions. These notes contain a succinct and cogent coverage of the material required for first ear undergraduate students in the Kenan Universities. It is also quite helpful to those students taking phsics related courses in other post secondar institutions like poltechnics. The tet could be emploed directl b teachers. Accordingl more time could be devoted to eperiments, discussions and problem solving. The notes provide the foundation for constructive revision prior to an eamination. Useful reminder of important details is given just before the worked eamples that preceed the problems that ever student should go through. SI units are emploed throughout. Outline Vectors, Rectilinear Motion, Projectile Motion, Forces & Newton s Laws of Motion, Circular Motion, Gravitation & Gravit, Simple Harmonic Motion, Elasticit, Surface Tension, Flow of liquids, Heat, Waves.

2 VECTORS Scalar and Vector Quantities Man quantities (e.g. volume) have no direction associated with them. These quantities which normall have magnitude onl are called scalar quantities. Scalar quantities include, mass, time, densit, work, temperature, amount of mone e. t. c. There are other quantities, which have both magnitude and direction. These are vector quantities. These include displacement, velocit, force, acceleration, and electric field e.t.c. A vector quantit is represented b an arrow draw to scale. The length of the arrow represents the magnitude and the direction of the arrow represents the direction of that vector. Vector Addition Vectors are added using the geometric method. Vectors don t obe ordinar rules of algebra. The combine according to certain rules of addition and multiplication. Vectors are added b geometricall connecting the head to the tail of the other vector and drawing a straight line between the other tail and head of the vectors. This gives ou a resultant vector AB + BC = AC = Resultant vector The resultant of a number of force vectors is that single vector which would have the same effect as all the original vectors together. Commutative Law of Vector Addition. AB + BC = BC + AB. During vector addition it does not matter with the vector ou begin with first. The resultant or effective vector will be the same. Associative Law of Vector Addition. Consider more than two vector which are to be added together. Draw to scale each vector in turn, taking them in an order of succession. The tail end of each vector is attached to the arrow end of the preceding one. The line drawn to complete the polgon is equal in magnitude to the resultant or equilibrant. An equilibrant of a number of vectors is that vector which would balance all original vectors taken, together. It is equal in magnitude but opposite in direction to the resultant. 1

3 For associative law of vector addition AB + BC + CD = AD, (AB + BC) + CD = AB + (BC +CD). Hence the ordering of the vectors makes no difference as far as their addition is concerned. This is the associative law of vector addition. Magnitude of Resultant Vector and Angles between the Vectors Consider the figure given below where AB v 1, BC v, AC v, DC v Sinθ, Angles CBD θ, CAB α, ACB β. AB +BC =AC, v 1 + v = v. To compute the magnitude of v we have (AC) = (AD) + (DC). But AD = AB + BD = v 1 + v Cosθ, DC = v Sinθ. Therefore (AC) = (v1+ v Cosθ) + (v Sinθ) = v 1 + v + v 1 v Cosθ or v= (v 1 +v + v 1 v Cos θ) 1/. To determine the angle we need onl find the angle α. From the figure we see that in triangle ACD, CD = AC Sinθ, and in triangle BDC, CD = BC sin θ; AC Sin α = BC Sinθ, v 1 Sin α = v Sinθ. Similarl, BE = v 1 Sin α = v Sin β which when we combine we get, ν ν = Sinθ Sinβ 1 = ν Sinα When v 1 & v are perpendicular θ = ½π and from v= (v 1 +v + v 1 v Cos θ) 1/ we have v = (ν 1 + ν ) ½ and tanα = (Opposite)/(Adjacent) = ν /ν 1 Subtraction of Vectors The negative of a vector is another vector of equal magnitude but opposite direction e.g.

4 The difference between two vectors is obtained b adding to the first the negative (or opposite) of the second. ν = v 1 - ν = (v 1 +( -v )). Note that v -v 1 = -v i.e. if the velocities are subtracted in the reverse order, the opposite vector results. Vector subtraction is anti-commutative. The magnitude of the difference is = v + v + v v Cos( π θ ) = v + v v v Cosθ. NB: The magnitude of a D vector quantit is basicall its length. Component of a Vector The component of a vector is its effective value in an given direction. e.g horizontal component of a vector is its effective value in a horizontal direction. A vector ma be considered as the resultant of two or more component vector. It s customar and most useful to resolve a vector into components along mutuall perpendicular directions. From the figure below we see that ν = ν + ν. But ν = νcosα, ν = νsinα. Defining unit vector ν & ν in the direction of the X and Y-ais, we note that v = OA = u v, v = OB = u v. Therefore v = u v + u v. In three dimensions we have v = u v + u v + u v. z z Multiplication of Vectors Operations of addition and subtraction can be carried out among like vectors. However in the case of vector multiplication, vectors of different kinds representing different phsical quantities can be multiplied, giving rise to another meaningful phsical quantit, e.g. F B = q o νb, where F B is the magnetic deflecting force in the magnetic field, ν is the drift velocit and B is the magnetic inductance. There are three kinds of operations for vector multiplication. (i) Vector Scalar = Vector (ii) Vector Vector = Scalar (iii) Vector vector = Vector (i) Multiplication of a vector with a scalar If a vector a is multiplied with an arbitrar number n (a scalar n), the resultant vector R will be n times the magnitude of a but the direction of R remains the same. n a = R = na Hence multiplication of a vector and a scalar gives a vector quantit in the same direction. 3

5 (ii) Scalar product (Dot. Product) Scalar product of two vector a and b represented b the smbol a.b (Read a dot b ) is defined as the scalar quantit obtained b finding the product of a and b and the cosine of the angle between the two vectors. a.b = ab cos θ a. a = a Since θ = 0. The product is zero for θ = π/. Hence the condition for perpendicularl is epressed b a.b = 0. The scalar product is commutative i.e., a. b = b. a since θ is the same in both cases. The scalar product is distributive with respect to the sum; i.e c (a+ b) = c. a + c. b. For unit vectors u, u, u z, we have u. u = u. u = u z.u z = 1, u. u = u. u z = u. u z = 0. Writing a and b in terms of their rectangular components, and appling the distributive law, we have a.b = ( u a + u a + u a ).( u b + u b + u b ) = a b + a b + a b. Note that a. a = a + a + a z z z z z (iii) Vector product (cross product) The cross product of two vectors as A and B (see figure below), written as A B = C defined as a vector C have magnitude C = C = ABSin θ and a direction of advance of a right hand screw when turned from A and B through angle θ, provided that the ais of the plane determined b A and B (the right - hand screw rule). Or if the curled fingers of the right-hand point from A and B the etended thumb points in the direction of C. Note that in accord with the right hand screw rule A B = -(B A) = C z z NOTE: (a) A B = 0, when A or B are either parallel or antiparallel. If vectors are parallel θ = 0 and the vector product is zero. The condition for parallelism is AB = 0. Obviousl AA = 0. (b) A B has its maimum when A and B are perpendicular to each other. (c) A B = -B A in contrast with A.B = B.A. Hence vector product anticommute (d) Null vector - has magnitude zero. (e) The vector product is distributive relative to the sum; i.e C (A + B) = C A + C B. The vector products among the unit vectors u, u, u z are u u = -u u = u z, u X u z = -u z u = u, u z X u = -u u z = u. Note that vector products among the unit vectors is cclic u u = u u = u z u z = 0 4

6 u u u z For A B = A A Az = B B Bz u (A B z - A z B ) + u (A z B - A B z ) + u z (A B - A B ) HAVE YOU READ AND UNDERSTOOD Scalar quantities? Vector quantities? Vector addition? Laws associated with vector addition? Vector resolutions & vector components? Vector subtraction? Vector multiplication? Summar of Ke Epressions. a.b =abcosθ a.b = b.a A B = - A B = AB Sinθ EXAMPLES E.1 Find the magnitude, component and direction w.r.t the -ais of the resultant vector formed b (a) A + B (b) A B if A = 5i +3j, B = i- 4j (a) Let P = A + B = (5+ )i + (3-4)j = 7i - j, P = (1+49) 1/ = 50 = 7.1. Components of the resultant are 7 and -1. Angle of the resultant tan 1 (P /P )= tan -1 (-1/7 ) =-8 0. The negative angle indicates that the angle is measured in the fourth quadrant. (b) Let P = A - B = 3i + 7j; magnitude = P = 7.6. Components of the resultant are 3 and 7. Angle of resultant with the X-ais is tan 1 (P /P )= tan -1 (7/3 ) = 67 0 E. If A + B + C = 0 and A = i + 3j + 4k and B = 5i+6j+7k then what is C and C. What is the angle between C and the X - ais. C = - (A +B) = - (7i + 9j + 11k); C = ( ) 1/ =(51) 1/ = Angle of the resultant with the X-ais is tan 1 (C /C )= tan -1 (9/7 ) = E.3 The vectors v 1 and v have same length v. The angle between them is θ. What is the magnitude of v - v 1?. 5

7 Let v = v - v 1. It is the base of the Isosceles triangle in the figure below 1 v In either right angled-triangle, Sin (θ/) =. Hence v = v Sin (θ/) v E.4 A Ferr - boat wishes to go straight across a stream that is flowing at 5km/h to the East. The pilot knows his speed w.r.t the water is 10km/h. At what angle must he head the boat and what will be his speed? Let be his vector velocit w.r.t. the water and v w be the velocit of the water. Then v' = v + v w - points to the north. V w - points to east. The triangle is right-angled triangle. Hence Sinθ = 1/, θ = Sin -1 (0.5) = The pilot must head the boat 30 0 to the west of north. The magnitude of v' is vcosθ = 10Cos30 0 = 8.66 km/hr. Note that this resultant vector is smaller in magnitude than the direct sum of its parts i.e E.5 A mass m is being pulled down on an incline plane b a gravitational force F g as shown in the figure. What is F the component of force that parallel to the surface? 6

8 Let X - ais be along or parallel to the surface. Drop a perpendicular from the tip of F g to the X ais. Then F = F g Sinθ E.6 A pilot of a private plane flies 0.0 km in a direction 60 0 north of east, then 30.0 km straight east, then 10.0 km straight north. How far and in what direction is she from the starting point? Let A = First displacement B = nd displacement C = 3 rd displacement R = Resultant A = 0 Cos 60 0 = 10 km; A = 0Sin60 0 = 17.3 km; B = 30Cosθ = 30 km; B = 30 Sinθ = 0 km C = 10 Cos 90 =0; C = 10 sin 90 = 10; R = ( ) = 40 Km; R = ( ) = 7.3 km Therefore R = R + R = = 48.4 The angle will be given b θ = tan -1 R 7.3 tan 1 = = R 40 E.7. Find the distance between the two points with coordinates (6,8,10) and (4,4,10) Since the Z- co-ordinate is the same, then both points lie in a plane parallel to the XY - plane. 7

9 R 1 = u (4-6)+u (4-8) + u z (10-10) = u (-) - u (4), r 1 = = 0 = 4.47 units E.8. Find the components of the vector that is 13 units long and makes an angle of.6 0 with Z-ais and whose projection in the XY plane makes an angle φ of 37 0 with the X-ais From V = V + V V = Vcos δ, V = V sin δ V = 13 units, θ =.6 0. Therefore Cos θ = 0.93 Z component = 13 cos θ = = 1 units Projection into XY plane V is given b V = V sin θ = 13 Sin.6 = = From which we have V = V cos φ = = 4 units; V = V sin φ = = 3.0 units. V = u (4) + u (3) + u z (1) or V = 4i + 3j + 1 k E.9 Find the angle between the vectors A = i +3j-k and B = -i+j+k Compute scalar product A.B = (-1) + 3(1) + (-1)= -1 A = 14 = 3.74, B = 6 =.45 From A.B =AB Cosθ, we can write Cos θ = θ = Cos -1 (-0.109) = A.B A B = = E.10 Find the vector product of the vectors A = i+3j - k and B = -i+j+k i AB = 1 j 3 1 k 1 = 7i-3j+5k PROBLEMS 1. Does a vector of zero length have a direction? 8

10 . Can a vector quantit ever have components different from zero but a magnitude of zero? 3. One sometimes speaks of the direction of time evolving from past to future. Does this mean that time is a vector quantit? 4. What does the term 'null' vector mean? 5. Calculate the components, magnitude and direction of the resultant vector P which is given b P = Q+ R, if Q = i + 3j and R = i + j. 6. Two vectors A and B have equal magnitudes and the angle between them is Calculate the magnitudes of the scalar product and the vector product of the vectors. 7. If A = i+3j, B = j-k, find A B and A B 8. In cartesian co-ordinate sstem, show that (i) i i = j j = k k =1 (ii) i i = j k = k k =0 (iii) ii = jj = kk =0 (iv) ij =k, jk=i, ki = j 9. Find the angle between two vectors A = 3i+4j+5k and B = 3i+4j-5k 10. Given the two dimension-less vectors A = 3i+4j and B = i+6j+5k determine (a) A+B (b) A-B ( c) B-A (d) B+A 11. Find the angle between a and b in the following cases: (a) a b = where a = and b = 3 (b) a b = -1/4 where a = 3 and b = 3 1. Find a.b, a.c and a.(b+c) given that (a) i +j+k = a (b) 3i+4j +5k = b (c) 4i+j-5k = c 13. Find the angle between the vectors a = i+k and b = 3i+4j+5k 14. Given a vector a in the -direction, a vector b in the -direction, and the scalar quantit d: (a) What is the direction ab? (b) What is the direction of ba? (c) What is the direction of (1/d)b? (d) What is the magnitude of a.b? 15. Show for an vector a, that a.a=a and that aa=0 16. Two vectors a and b have components, in arbitrar units, a =3., a =1.6; b =0.50.b =4.5 (a) Using the unit vector i and j, epress a and b. (b) Find the angle between a and b, (c) Find the components of a vector c that is perpendicular to a in the XY-plane and has a magnitude of 5.0 units. 17. (a) Define scalar and vector quantities. (b) When is the sum of the two vectors maimum and when is it minimum? (c) Can two vectors of different magnitude be added in such a wa that their resultant is zero? (d) Two vectors have non-zero magnitude. Under what condition will their dot and cross product be zero? (e) If A and B are arbitrar vectors and the are such that A.(B A) =0. Eplain. 18. The resultant of two vectors is 30 units long and forms with their angles of 5 0 and Find the magnitude of the two vectors. 19. Find the angle between two vectors 8 and 10 units long, when the resultant vector makes an angle of 50 0 with the larger vector. Also calculate the magnitude of the resultant vector. 0. Two vectors, 10 and 8 units long form an angle of 60 0, 90 0 and Find the magnitude of the difference and the angle with respect to the larger vector. 9

11 1. Find the rectangular components of a vector 15 units long when it forms an angle w.r.t. the positive X-ais of 50 0, 130 0, 30 0 and Given for coplanar vectors 8, 1, 10 and 6 units long respectivel; the last three make with the first angles of 70 0, and 00 0 respectivel. Find the magnitude and direction of the resultant vector. 3. Prove that if the sum and the difference of two vectors are perpendicular, then the vectors have equal magnitudes. 4. Prove that if the magnitudes of the sum and the difference of two vectors are equal the vectors are perpendicular. 5. Epress the triple product V 1.(V V 3 ) in determinant form. Derive, from it its smmetr properties i.e. V 1.V V 3 = V 3.V 1 V = V.V 3 V 1 Prove that the value of the triple product is equal to the volume of the 11 parallelepiped made from the three vectors 10

12 RECTILINEAR MOTION Mechanics deals with the relations of force, matter and motion. A mathematical method for describing motion is a branch of mechanics called Kinematics. Kinematics is the stud of motion without reference to the forces, which ma cause that motion. Motion is basicall a continuous change of position. The simplest case of motion is that of a point along a straight line. The motion of a bod is rectilinear when its trajector is a straight line. Distance This is a measure of length between two points not necessaril in a straight line. It s a scalar quantit distance is measured in meters. Displacement This is the shortest measure of length between two points in a specific direction. It s linear distance in a given direction. It s a vector quantit. Its SI units are meters (m) Speed This is the time rate of change of distance i.e. it is a scalar quantit measured in m/s. The average speed ν of a bod, which travels a distance s in time t, is defined b v =ds/t. From which ds = vt. Velocit This is the time rate of change of displacement. It is a vector quantit and its SI units are m/s. The velocit of a bod changes if there are changes in magnitude or in direction of motion or in both. The average velocit during a certain time interval is equal to the average displacement(s) per unit time during that time interval. Therefore v average = S/ t. A bod is said to have move in uniform velocit when S/ t = constant hence its acceleration is zero i.e. velocit is constant. Displacement - Time Graph The gradient at an instant or at a given time of a displacement-time graph for a bod that is changing positions represents the instantaneous velocities. If it is a straight-line graph then it is undergoing uniform velocit otherwise it is non-uniform velocit. Negative gradient means that the bod is moving back. Increasing gradient is increasing velocit and vice versa. The are under the curve has no phsical meaning. We can obtain the instantaneous velocit b comparing the time derivative of the displacement, v = ds/dt. Acceleration This is the time rate of change of velocit. It s a vector quantit with SI units as m/s. a = dν/dt (also equivalent to acceleration). A bod is said to move in uniform acceleration when ν/ t = constant. i.e. the velocit changes b equal amounts in equal times. Non-uniform acceleration is due to the different changes in velocities in equal time intervals. Velocit-Time Graph The gradient at a given point of a velocit - time graph gives the instantaneous acceleration at that point. If the graph is linear then the bod is undergoing uniform acceleration otherwise it is nonuniform acceleration. Increasing gradient means increasing acceleration and vice versa. Negative gradient implies negative acceleration or declaration. The area under the curves gives the 11

13 displacement covered. Area under curve = υ. dt = displacement. Area above time ais is positive and below it is negative meaning negative displacement or backward motion. Equations of Motion If a bod is moving in a straight line under constant acceleration, then relations among its velocit displacement, time and accelerations can be represented b equations. These equations are called equations of motion. Suppose a bod starts with an initial velocit u, and has a constant acceleration a. Suppose it covers a distance (displacement) in time t and its velocit becomes ν. The constant acceleration a is given b a = (ν - u)/t which on rearranging gives ν = u + at. This is the first equation of motion. Since the acceleration is constant, the average velocit ν of the bod is ν = ½(u + ν). The displacement s covered in time t is s = νt or s = ½( u + ν) t. Using the first equation of motion to eliminate ν, we have s = ½ t (u + u + at) = ½ t(u +at). Therefore s = ut + ½(at ). This is the second equation of motion. The third equation of motion can be obtained b combining the first two equations of motion. Square both sides of equations ν = u+at v = (u+at) = u +uat+ a t = u + a (ut + ½(at )) But s = ut + ½(at ). Hence v = u + as. This is the third equation of motion. Therefore the three equations of motion are (i) v = u + at (ii) s = ut + ½(at ) (iii) v = u + as These equations hold well onl when the acceleration is constant and the motion is in a straight line. If we know an three of v, u, a, s and t, then the remaining parameter can be evaluated with ease. The three equations can also be found b integration. v dv = a dt dv = a dt u dv dt t 0 ds dt. [ ] [ ] t u at o dv ds dv ds v v = v -u = at. Therefore v = u + at. s v s 1 0 = 0 u Also a = =. = v. a. ds = v. dv [ as] Hence v = u + as. When the bod start from rest the initial velocit u =0 and we get v = at, s =½(at ), v = as. v v u Equation of Motion under Gravit The earth attracts everbod towards its center. As a result of this attractions a constant acceleration is produced in bodies falling freel towards the earth. This is called the acceleration due to gravit and is denoted b g. The motion of bodies falling towards the earth (or thrown awa from the earth) is called gravitational motion. Air resistance is neglected. If in the three equations we replace a b g, s b h then we shall get equations for the bodies falling towards the earth. Thus these equations are ν= u +gt, h = ut + ½(gt ), ν = u + gh, where h is the height from the earth's surface. g is negative for bodies moving upwards and is positive for bodies moving downwards. If initial velocit is zero, then ν =gt, h =½(gt ), ν =gh. For free-fall motion, the particle ma have zero velocit, et it can be accelerating. HAVE YOU READ AND UNDERSTOOD The difference between (i) distance and displacement (ii) Speed and velocit? Uniform (i) velocit? (ii) Acceleration? Equations of motion and its applications? 1

14 Summar of Ke Epressions v = ds/dt v = u + at s = ut + ½ at v = u + as EXAMPLES. E.1 The velocit of a car is retarded from 10m/s to 4m/s in seconds. What is its acceleration? We have been given v = 4, u = 10, u =, a =? For retardation a is negative. Using v = u + at a = (u - v)/t = (10-4)/ = 3 m/s (retardation) E. An object is dropped into a well and hits the water seconds after being released. How deep is the well? Take g=10m/s We have been given u = 0, t =, g = 10, h =? h = ut + ½ at = / 10 4 = 0 m E.3 A ball is thrown verticall into the air at 50m/s. How high will it rise and how long will it take to reach that height? G = 10m/s We have been given u = 50m/s, v = 0, g = 10m/s, h =?, t =?. g is negative since it is against gravit. v = u - gh. h = ( u - v )/g = u /g = (50 50)/( 10) = 15mm From v = u - gt, t = (u - v)/g = u/g = 5 seconds E.4 A particle is fired with a constant velocit of m/s into a region where it is subjected to an acceleration of 10 1 m/s directed opposite to the initial velocit. How far does the particle travel before coming to rest? Ho long does the particle remain at rest. We have been given u = m/s, v = 0, a = m/s. Using v = u + as we have S = (v - u )/a =( )/(10 1 ) = 0.5m Also t = (v-u)/a = ( )/( ) = Sec = 0.5µsec. E.5 A coin is thrown verticall upwards from the ground with a speed of 10m/s. (a) How long does it take to reach the highest point? (b) What is the maimum height reached b the coin? 13

15 We have been given a = g = 10m/s, u = 10m/s; at the heighest point v = 0; t = (v - u)/g = -10/-10 = 1 sec Ma height = ut - ½gt = ½10t = 5.0 m E.6 Describe the motion represented b figures below. (i) O Stationar O-A Acceleration (increasing slope) A-B Moving with constant velocit B-C Decelerating (decreasing slope) C-D Stationar D-E Accelerating and moving back to the starting point E-F Moving with constant velocit F Stationar at starting point (ii) A-B B-C C-D D-E E-F At F F-G G-H H Stationar & awa from origin Accelerating back to the origin Constant velocit Moving awa from the origin with constant velocit in the opposite direction to the original direction Decelerating. Momentaril at rest. Accelerating back toward origin Constant velocit towards origin Stationar at origin PROBLEMS 1. Define distance, speed, displacement, velocit, uniform velocit, and uniform acceleration. In each case eplain whether it is a vector or scalar quantit. State their SI units.. An automobile travels on a straight road for 40km at 30 km/h. It then continues in the same direction for another 40km at 60km/h. What is the average speed of the car during the 80km trip? Sketch the motion (on a displacement - time) of the automobile for the whole trip giving the relevant distance and time. 3. A car accelerates at 9.km/h. Compute the acceleration in m/s. 4. A particle had a velocit of 18m/s and.4s later its velocit was 30m/s in the opposite direction. What was the average acceleration of the particle during this.4s interval? 14

16 5. On a dr road a car with good tires ma be able to brake with a deceleration of 4.9m/s. How long does such a car initiall traveling at 4.6m/s take to come to rest? How far does it travel in this time? 6. Can an object (i) have zero velocit and still be accelerating? (ii) have a constant speed and still have a varing velocit? In each case give an eample if our answer is es; eplain wh if our answer is no. 7. A rocket ship in free space moves with constant acceleration equal to 9.8m/s. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light? (b) How far will it travel in so doing? (Speed of light = m/s). 8. Derive the three equations of motion in a straight line for a general case and for a bod starting motion from rest. Write down equations of motion under gravit (Assume acceleration is constant). 9. A car runs at a constant speed of 15m/s for 300 seconds and then accelerates uniforml to a speed of 5m/s over a period of 0 seconds. This speed is maintained for 300 seconds and then accelerates uniforml in 30 sec. Draw a velocit time graph to represent the journe described above. From the graph (i) find the acceleration while the bod changes from 15m/s to 5m/s. (ii) the total distance traveled in the time described. (iii) The average speed over the time described. 10. An object moves in a straight line starting from rest. There are two stages in the motion (a) It gains speed uniforml for seconds and attains a speed of 6.0 m/s (b) It continues with this speed for a further 1.5 seconds. Find the acceleration in both stages (a) and (b) and the total distance moved during both stages. 11. A stone is thrown verticall upward from the ground and is observed to pass a point situated at a height H above ground level at 4 sec and gain at 5 seconds after projection. Find the velocit of projection of the stone and the height of H. 1. A bod slides down a frictionless incline, which makes an angle of 30 0 with the horizontal. Compute the velocit v after sliding 8 metres from rest and the time t it takes to slide the 8m. 13. A bod is projected up a smooth 30 0 incline with a velocit of 40m/s measured along the slope. Determine (a) the time it takes to return to the starting point (b) the distance it moves along the slope before reaching its highest point. 14. Prove that a gun will shoot three times as high when its angle of elevation is 60 0 as when it is 30 0, but will carr the same horizontal distance. 15. A ball after having fallen from rest under the influence of gravit for 6 seconds crashes through a horizontal glass plate, thereb loosing /3 of its velocit. If it then reaches the ground in seconds find the height of the plate above the ground. 16. A bead (fig. shown) is free to slide down a smooth wire tightl stretched between points P 1 and P on a vertical circle. Find (a) its velocit v on arriving at P and (b) the time taken to arrive at P and show that this time is the same for an chord drawn from P. 15

17 Bead acc. along the wire = gcosθ; length of wire = Rcosθ 17. Bod 1 is released from rest at the top of a smooth inclined plane and at the same instant bod is projected upward from the foot of the plane with such velocit that the meet halfwa up the plane. Determine the velocit of projection and the velocit of each bod when the meet. 16