DEPARTMENT OF MECHANICAL ENGINEERINGSAVEETHA SCHOOL OF ENGINEERINGSAVEETHAUNIVERSITY, CHENNAI 602 105ACADEMIC YEAR 2014- 2015

QUESTION BANKSub. Code/Name: MEX101-Vector Mechanics Year/ Sem: I1. What is vector Mechanics?It is a branch of science that deals with the behavior of the body, when it is subjected to the action of forces in rest or in motion.2. Differentiate kinetics and kinematics? In kinetics both the motion and its causes are considered. Kinetics relates the action of forces and the resulting motion. Kinematics is concerned with the description of motion of objects independent of causes of motion. Hence here the study is made of motion in the inter relation among position, velocity, acceleration and time without taking into the forces causing motion.3. Define a Particle?It is a small amount of matter, which may be assumed to occupy a single point in space. Its dimension is considered to be negligible but it has mass.4. Define a Rigid Body?It is a combination of a large number of particles occupying fixed positions with respect to each other. 5. Define Vectors?Vectors or Vector quantities are defined as those physical quantities, which have both magnitude and direction and obey the parallelogram law of addition of vectors.6. Classify the vectors. Vectors are classified as, i) Free vector ii) Sliding vector iii) Fixed or (bound) vector iv) Unit vector v) Negative vector vi) Null or (zero) vector7. Mention the Vectors Operations? Vector Addition Vector Subtraction Cross product of vectors Dot product of vectors 8. If =2i+4j-7k and =3i-5j+6k, find (i) 2 - (ii) . 5 Solution:(i) 2= 2(2i+4j-7k) = (4i+8j-14k) 2 - = (4i+8j-14k) (3i-5j+6k) = (i+13j-10k)(ii) 5 =5(3i-5j+6k) = (15i-25j+30k) . 5 = (2i+4j-7k).(15i-25j+30k) = (2x15) i (4x25) j (7x30) k = (30i 100j - 210k)9. Define force. List the characteristics of a force.Force is an external agent which changes (or) tends to change the state of rest (or) of uniform motion of a body upon which it acts. The characteristics of a force are Magnitude, Line of Action, and Direction.10. On what fact a force system is classified? A force system is classified on the fact whether all the forces lies on the same plane or not, direction and line of action of the forces.

11. What are coplanar forces?Coplanar forces mean the forces will acts in the same plane. 12. Define concurrent force system?The line of action of all the forces will passes through a single point in a system of force is called as Concurrent force system.13. What do you mean by resultant force?The number of forces acting on a particle is replaced by a single equivalent force, which produces the same effect as that of all the given forces is called as Resultant force.14. What is resolution of force?Splitting up a force into components along the fixed reference axes is called resolution of a force.

15. State principle of transmissibility of forces.It states that If a force acts at any point on a rigid body it may also be considered to act at any other point on its line of action.For example consider a force P acting on a rigid body as shown in figure below. Here the two forces P and P have the same effect on the rigid body and are said to be equivalent.

16. State triangle law of force?It states that When two forces acting upon a particle represented both in magnitude and direction by the two sides of triangle taken in order, then the resultant will be represented by the third side of the triangle in the opposite order

17. State polygon law of forces?This law states that, When a number of forces acting on a body represented in magnitude and direction by the sides of an open polygon taken in order, the closing side of the polygon taken in opposite order represents the resultant in magnitude and direction18. State the Parallelogram Law of Forces?It states that, If two forces acting on a particle are represented in magnitude and direction by two adjacent sides of a parallelogram, then the resultant of these two sides is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.19. List out the methods of finding the resultant force?Using the following three laws we can find the resultant force by graphical method and analytical method ;( i.e. Triangle law, Parallelogram law, Polygon law)20. Two concurrent forces of 12N and 18N are acting at an angle of 60. Find the resultant force. Let P=12N, Q=18N, =60 Resultant force, R= R= = 26.15NInclination of resultant force with the force P,tan = = =0.742 = tan -1(0.742) = 36 34PART B (16 Mark Questions)1. Four forces are acting on a particle. The magnitude of the forces are 150 N, 200 N, 80 N, 180 N and their respective angles with the horizontal are 30, 150, 240, and 315. Calculate the magnitude and direction of the resultant. Assuming that all the forces are acting away from the point.2. Determine the x & y components of each of the forces shown in Fig. 1.1 and also find the resultant and the directions?

Fig. 1.1

3. Find the magnitude and direction of the resultant from the following cases.a. 30 N inclined at 250 towards North of Eastb. 25 N towards Northc. 35 N inclined at 400 towards South of Westd. 30 N towards North West. 4. A car is pulled by means of two cars as shown in figure 1.2. If the resultant of the two forces acting on the car A is 40 KN being directed along the positive direction of X axis, determine the angle of the cable attached to the car at B , such that the force in cable AB is minimum. What is the magnitude of force in each cable, when this occurs?

Fig 1.25. The force system shown in the Fig. 1.3 has resultant of 520 N along the negative direction of y-axis. Compute the values of force P and .

260N1252343200NR=520N

360NP

Fig. 1.36. Find the resultant of space concurrent force as shown in Fig. 1.4.

Fig. 1.47. Determine the x and y components of each of the force shown in Fig. 1.5 and also find resultant and directions? 120mm

35mm105mm100 mm115 mm60 mm250 N145 N

225 Nm

Fig. 1.58. If the five forces act on a particle as shown in Fig. 1.6 and if the algebraic sum of horizontal components of all these forces is -300 KN. Calculate the magnitude of F and the resultant of all the forces. YX150kN100kN250kN75kNF45o3430o

Fig. 1.6

UNIT II - EQUILIBRIUM OF PARTICLES AND RIGID BODIESPART A (2Marks)1. State Newtons First law of motion.If the resultant force acting on a particle is zero, then the particle will remains at rest (or) will move with constant speed in a straight line2. What is equilibrium of forces?Equilibrium of a Particle is defined as, When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium. 3. Define Lamis Theorem?When the forces acting at a point are in equilibrium, then the magnitude of the each force will be proportional to the sine of the angle between the other two forces.

Let the forces acting at point O be P, Q, R then according to Lamis Theorem

4. Define Moment.Moment of a force (or) axis is a measure of the tendency of the force to rotate a body about that point or axis.5. Define Equilibrant?It is defined as the force with the magnitude equal to that of the resultant but opposite in direction and this force brings the body to rest from its motion.6. State Varignons theorem?Therefore Varignons Theorem states that the moment about a given point O of the resultant of several forces is equal to the sum of the various moments of the various forces about the same point O7. What is meant by free body diagram?The sketch showing all the forces and moments acting on the body which is isolated from the surrounding bodies is called as free body diagram.8. Differentiate action and reaction of forces?Consider a ball placed on a horizontal surface, the self-weight of the ball is acting vertically downwards through its Centre of gravity, and this force is called Action. Now the ball can move horizontally but its vertical downward motion is resisted due to resisting force developed at support its acting vertically upwards, this force is called Reaction.9. What are the various types of supports?Important Types of supports are Roller support Simple or Knife edge support Fixed support Hinged or Pin joint support10. List out the different types of beamsThe various types of beams are, Simply supported beam Cantilever beam Continuous beam Overhanging beam Fixed beam11. What are different types of loads acting on beams?1. Concentrated Load2. Uniformly Distributed Load (Udl)3. Uniformly Variable Load (Uvl)12. Define Concentrated Loads?

This load is also called as Point load. The above figure shows two loads P1& P2 acting at points L & M respectively on a simply supported beam.13. Define Uniformly Distributed loads?

When the load w per unit length has a constant value along the length of the beam, the load is called as the uniformly distributed load.14. Define uniformly varying load?

In this type of loading, the beam carries load in such a way that the rate of loading on each unit length of the beam varies uniformly.15. Differentiate uniformly distributed load and uniformly varying load?When the load w per unit length has a constant value along the length of the beam, the load is called as the uniformly distributed load. In UVL type of loading, the beam carries load in such a way that the rate of loading on each unit length of the beam varies uniformly.16. State the necessary and sufficient conditions for equilibrium of a particle in plane?H = 0, V = 0, R=017. Define Couple?The moment created by two equal, opposite and non-collinear force is known as couple.18. Examine the equilibrium condition for A shown in below diagram?

Solution: Resolving horizontally,Fx =150 - 200 cos60 100 cos60 = 0Resolving vertically,Fy = 200 sin60 - 100 sin60 - 86.6 = 0As Fx & Fy are zero, A is under equilibrium.19. What is the condition of equilibrium of a particle in space?Equilibrium of a particle in Space: A particle A is in equilibrium if the resultant of all forces acting on A is zero. The components Rx, Ry, Rz of the resultant are given by the relationRx = Fx ; Ry = Fy ; Rz = Fz ; Expressing that the components of the resultant is zero, we write Fx = 0; Fy = 0; Fz = 0The above equation represents the necessary and sufficient conditions for the equilibrium of a particle in space. 20. What is force couple system?PART B (16 Marks)

1. Determine the tension in cable PQ and PR required to hold 40 kg weight shown in Fig. 2.1.

Fig. 2.12. A string ABCD, attached to two fixed points A and D has two equal weight of 1000N attached to it at B and C. The weight rest with the portions AB and CD inclined at angle of 300 and 600 respectively ,to the vertical as in fig 2.Find the tensions in the portion AB,BC and CD of the string, if the inclined of the portion BC with the vertical is 1200

3. Two identical rollers, each of weight 50 N, are supported by an inclined plane and a vertical wall as shown in fig 3.Find the reaction at the point of supports A,B and C. Assume all the surface to be smooth.

4. Two roller, each of weight 50N and of radius 10cm rest in horizontal channel of width 36cm as shown in fig4. Find the reaction on the point of contact A,B and C.

5. ABCD is a weightless rod under the action of four forces P,Q,S and T as shown in fig 5. If P=10N,Q=4N,S=8N and T=12N, calculate the resultant and mark the same in direction with respect to the end A of the rod.

6. Four forces of magnitude and direction acting on a square A, B, C, D of side 2 m are shown in Fig. 2.6, calculate the resultant in magnitude and direction and also locate it point of application with respect to the sides AB and AD.

Fig. 2.67. Find the reactions at support A and B. Distance between A & C, C & D is 2 m, distance between D & B is 4 m and distance between B and E is 3m which is shown in Fig. 3.1.ACDBE10 kN25 kNm5 kN/m10 kN/m20 kN30

Fig. 2.7

8. The three forces and a couple of magnitude, M = 18 Nm are applied to an angled bracket as shown in figure 2.8. A) find the resultant of this system of forces. B) locate the points where the line of action of the resultant intersects line AB and BC .

Fig 2.8.

UNIT III CENTROID AND MOMENT OF INERTIAPART A (Two Mark Questions)1. Define centroid and centroidal axis?It is the axis which passes through the centroid of an area.It is the commonly used axis.

2. Define Centre of GravityThe centre of mass (gravity) of a body or a system of particles is the resultant of the weights of the individual particles making up the body or system. The centre of gravity of a collection of masses is the point where all the weight of the object can be considered to be concentrated. 3. Define Centre of massIt is the point at which the sum of the masses of all individual particles is concentrated.If x, y and z are the coordinates of the centre of mass then we have,If x, y and z are the coordinates of the centre of mass then we have,

, ,

4. Distinguish between centroid and centre of gravity.

5. Define Parallel-axis theorem?When the second moment of an area has been determined with respect to a given axis, the second moment of area with respect to a parallel axis can be obtained by parallel axis theorem. (Also known as transfer formula). As per parallel axis theorem we can write, IAA = Ixx + A ( hA)2 IBB = Ixx + A ( hB)2 Moment of inertia of area about an axis AA or BB is greater than the moment of inertia of an area about parallel centroidal axis by a positive amount Ah2 where, h is the distance between the centroidal axis and the given axis.5. Define Perpendicular-axis theorem?Moment of inertia of a plane lamina about an axis perpendicular to the lamina and passing through its centroid is equal to the sum of the moment of inertias of the lamina about two mutually perpendicular axes passing through the centroid and in the plane of the lamina.6. Define radius of gyration

7. What is the equation for first and second moment of area A about the X- axis?The first moment of area A about the x-axis as Mx = y dA. The equation for second moments is given by Ixx = y2 dA, Iyy = x2 dA

8. Define Pappus Guldinus Theorem :( with respect to area)?The area of a surface of revolution is equal to the product of the length of the generating curve and the distance travelled by the centroid of the curve while the surface is being generated9. Define Pappus Guldinus Theorem :( with respect to volume)?The volume of a body of revolution is equal to the product of the generating area and the distance travelled by the centroid of the area while the body is being generated.

10. Define moment of inertia of a body?It is defined as the quantitative estimate of the relative distribution of area and mass of a body with respect to some reference axis.11. Write down the equation of moment of inertia about centroidal axis Ixx, Iyy of a rectangle?Ans:

bd

12. Write down the equation of moment of inertia about centroidal axis Ixx, Iyy of a right angled triangle?hb

Ans:

2030

13. Find out the centroid of given figure?

14. Write down the equation of moment of inertia about centroidal axis Ixx, Iyy of a circle?

15. Write down the equation of moment of inertia about centroidal axis Ixx, Iyy of a semicircle?

16. Write down the equation of moment of inertia about centroidal axis Ixx, Iyy of quadrant of circle?

2020

17. Find out the centroid of the given figure

PART B (16 Marks)1. Find the centroid of I section with top flange 100 mm x 20 mm, web 200 mm x 30 mm and bottom flange 300 mm x 40 mm.2. Locate the centroid of the shaded area shown in the figure 3.1.

Fig. 3.13. The Fig. 3.2 shows the dimensions 10 x 10 x 2 cm. Determine the moment of inertia of the section about the horizontal and vertical axes, passing through the centre of gravity of the section?

Fig. 3.2

4. Find the moment of inertia of the section shown in Fig. 3.3 about the centroidal axis X-X perpendicular to the web?

Fig. 3.3

5. Find the centroid of the channel section in Fig. 3.4 (All the dimensions are in mm)10020105010

Fig. 3.4

6. Determine the moment of inertia for the following section at centroidal axis, which is shown in Fig. 3.5. All dimensions in cm?505040202020

Fig. 3.5

7. Find the moment of inertia of the section shown in figure about its horizontal centroidal axis. (All dimensions are in mm)

Fig. 3.68. Calculate the x and y co-ordinate of the centroid of the shaded area is shown in Fig. 3.7.

9. Fig. 3.7

Fig. 3.7

Unit IV- FRICTIONPart A (2marks Questions and Answers)1. What is a Frictional force? A force is exerted at the surface of contact by the stationary body on the moving body when it moves on it. This force is called the frictional force. The frictional force always acts in a direction opposite to the direction of motion.2. Define Friction?Friction can be defined as the property of the bodies by virtue of which a force is exerted by the stationary body on the moving body, to resist the motion of the moving body.3. What is Fluid Friction? This force is exerted between the layers of fluid moving at different velocities. In this case the surfaces have thick layer of lubricant .It is also called as film or viscous lubrication. If the rubbing surfaces have thin layer of lubricant it is called as boundary or non-viscous lubrication. It is the basic in the analysis of lubricated mechanisms.4. What is dry friction?This type of force is exerted between non-lubricated surfaces or dry surfaces of the bodies at the surface of contact. Dry Friction describes the tangential component of the contact force existing between two dry surfaces, which slide or tend to slide relative to one another. Dry friction is also called as coulomb friction. Dry friction is of two types: a) Static friction and b) dynamic friction.5. What are the various types of friction?1) Dry friction2) Fluid friction3) Internal friction6. Define Limiting friction?It is the maximum frictional force exerted at the time of impending motion, i.e when the motion is about to begin.7. What are the causes for the different types of resistances by a wheel? The causes are due to1) combinmed effect of axle friction and friction at the rim2) the deformation of the wheel and the ground when it is in contact not at a single point but over a certain area.8. What is belt friction?The belts or ropes are used to transmit power from one shaft to another by means of pulleys which rotate at the same speed or at different speeds. A belt or rope drive is a device with belt and pulley arrangement in which the belt friction is used for: raising a load, transmitting power or application of brakes to stop motion.9. What is the cause for coulomb friction?The main cause for this type of friction to occur is due to the interlocking of the minute projections on the surfaces which oppose the relative motion.10. What are the two types of dynamic friction?1) Sliding friction2) Rolling friction

11. Define Angle of friction?It is defined as the angle of inclined plane at which a block or body begins to slide down the plane. It is also called as the angle of repose.12. What are the advantages of Friction? Advantages of Friction:a. Friction enables us to walk b. It enables to transmit power from one machine to another through a belt.c. Screws and nails hold the board together due to friction.d. Friction plays an important in the application of brake.13. What are the disadvantages of Friction?Disadvantages:a. It leads to wear and tear in machinesb. It reduces the efficiency of the enginesc. Large amount of power is lost to overcome frictiond. It causes heat between the surfaces

14. What are the methods using to reduce friction in machines?Methods for reducing friction in machines:a. Use of lubricants between the surfacesb. Use of antifriction metalsc. Polishing the surfaces that come in contactd. Use of roller bearings and ball bearings16. State the law of Coulomb frictions?a) the direction of frictional force on a surface is such as to oppose the tendency of one surface to slide relative to the other.b) the total frictional force is independent of the area of contact between two surfaces and depends on the nature (smoothness/roughness) of surfaces in contact.c) the magnitude of frictional force is exactly equal to the force which tends to move the body till the limiting value is reached.d) the magnitude of limiting friction bears a constant ratio to the normal reaction between the two surfaces.e) Friction force is never greater than the force required to prevent motion.PART B (16 MARKS)1. Block (2) rests on block (1) and is attached by a horizontal rope AB to the wall as shown in Fig. 4.1. What force P is necessary to cause motion of block (1) to impend? The co-efficient of friction between the blocks is and between the floor and block (1) is 1/3. Mass of blocks (1) and (2) are 14 kg and 9 kg respectively.

Fig. 4.12. A block A weighing 2000 N resisting on a horizontal surface supports a block B weighing 1000 N as shown in Fig. 4.2. The block B is connected to a string which is attached to a vertical wall. Find the horizontal force P which should be applied on the block A so as to adjust move it leftwards. Take = 0.3 at all contact surfaces. Find also the corresponding tension of the string.

Fig. 4.23. Block of weight W1=1290 N rests on a horizontal surface and supports another block of weight W2 =570 N on top of it as shown in Fig. 4.3. Block of weight W2 attached to a vertical wall by an inclined string AB. Find the force P applied to the lower block that will be necessary to cause the slipping to impend. Coefficient of friction between block 1 & 2 is 0.25. Coefficient of friction between block 1 and horizontal surface is 0.4.

Fig. 4.34. Two blocks A of weight 48 N and B of weight 80 N are on rough horizontal surface as shown in Fig. 4.4. Find the minimum value of P to just move the system. If the coefficient of friction between block A and ground is 0.3 and that between block B and ground is 0.25. And also find the tension in the string.

Fig. 4.45. A body of weight 500 N is lying on a rough plane inclined at an angle of 250 with the horizontal. It is supported by an effort (P) parallel to the plane as shown in Fig. 4.5. Determine the minimum and the maximum values of P, for which the equilibrium can exist, if the angle of friction is 200.

Fig. 4.5

6. Two masses m1 and m2 are tied together by a rope Parallel to the inclined plane surfaces, as shown in Fig. 4.6. Their masses are 30 kg and 10 kg respectively. The coefficient of friction between m1 and the plane is 0.25, while that of mass m2 and the plane is 0.5. Determine (i) the value of the inclination of the plane surface for which masses will just start sliding, (ii) the tension in the rope.

Fig. 4.6

7. A 6 m ladder weighing 400 N rests against a smooth wall is shown in Fig. 4.7. The angle between it and the floor is 70o. The co-efficient of friction between the floor and the ladder is 0.25. How far the ladder can a man of 80 kg walk before the ladder slips?

Fig. 4.7

8. Determine the horizontal force P required for wedge B to raise the block A of weight 4500 N is shown in Fig. 4.8, if the coefficient of friction on all surfaces is 0.2.

Fig. 4.8

Unit V DYNAMICS OF PARTICLESPart A (2marksQuestions and Answers)1. Define displacement?Displacement can be defined as the change in the position of a particle i.e. the distance moved by the particle with respect to certain fixed point.Displacement can be expressed asx = x x0where x = position of particle at time t sec x0 = position of particle at time 0 sec

2. Define average velocity?Average velocity of the particle is defined as the ratio of the displacement x and time interval t.Let the position at time t = xLet the position at time t + t = x + x

3. Define relative motion?Relative motion is defined as the motion of the body that may be evaluated with respect to another moving body.4. State Newtons First Law of motion?First Law: According to this law Every particle continues in its state of rest or of uniform motion in a straight line unless it is acted upon by some external force. This law is also called as Law of Inertia5. State the principle of conservation of momentum?It states The total momentum of a system of masses in any one direction remains constant, unless it is acted upon by an external force in that direction. It is applied to problems on impact. If the resultant impulse acting along a particular direction is zero for a body or system of bodies, then momentum is conserved (unchanged) along that direction.6. Define impact?Impact: It is a collision between two bodies, which occurs in a very short interval of time during which the two bodies exert relatively large forces on each other.

7. Define Line of impact?Line of Impact: The common normal to the surfaces of two bodies in contact during impact is called line of impact.

8. Define Central impact?Central Impact: when the mass centers of colliding bodies are located on the line of impact, it is called central impact.

9. Define Period of restitution? Period of restitution is the period of time from the end of compression stage to the instant when the bodies separate.10. Define period of deformation?Period of deformation is the time taken by the bodies from the instant of initial contact to the instant of maximum deformation.

11. Define coefficient of restitution?The ratio of the relative velocity after impact to the relative velocity before impact is called as Coefficient of Restitution and it is represented by e.

12. What do you mean by perfectly inelastic body?When coefficient of restitution is equal to zero then the bodies are perfectly inelastic.

13. What do you mean by perfectly elastic body?When coefficient of restitution is equal to one then the bodies are perfectly elastic.14. What do you mean by acceleration and average acceleration?Acceleration: It is defined as the rate of change of velocity of a particle with respect to time. Acceleration is a vector quantity.

Average acceleration: Acceleration of a particle over the time interval t is defined as the quotient of t and v: The unit of acceleration is expressed as m/s2.15. What do you mean by Instantaneous acceleration?Instantaneous acceleration: Acceleration of the particle at the instant t is obtained from the average acceleration by choosing smaller and smaller values for t and v:

The unit of acceleration is expressed as m/s2.16. Define translation and rotation?If an imaginary straight line drawn on the body remains parallel to its original position during the motion, then the body is said to have a translatory motion.When all the particles of a rigid body move in concentric circular paths, then the body is said to have a rotational motion.

17. Define D-Alemberts Principle?The Force system consisting of external force and inertia force can be considered to keep the particle in equilibrium. Since the resultant force externally acting on the particle is not zero, the particle is said to be in dynamic equilibrium. This principle is known as D-Alemberts principle.18. Differentiate Rectilinear Motion and Curvilinear Motion?The simplest motion along a straight line and such motion is known as rectilinear motion. The important parameters required to study this motion along a straight line are;- position, velocity, displacement, acceleration, distance and speed.Curvilinear Motion: When a particle moves along a curved path, the motion is said to be curvilinear motion. When this motion is confined to one plane, the motion is said to be plane curvilinear motion.19. Write down the equations of motions?Equations of Motions:1)

2)

3)

4)

Where v = Final velocityu = Initial velocitya = accelerationx = displacement in timet seconds

Distance travelled in the nth second When the body falls from certain height say h,then

Final velocity = 20. Define work-energy Principle?When a body is acted upon by a force, then the work done by the resultant force is equal to the change in kinetic energy of the body.21. What do you mean by Impulse and Impulsive force?Impulse is defined as the product of force acting on the body and the time for which it acts.Impulse = Force timeI = F tImpulsive Force:It is defined as the force, which acts for a very short time. It is also called as Blow. It occurs in collisions, explosions etc.

PART B (16 marks)1. A body weighing 200 N is pushing up a 30o plane by force of 400 N acting parallel to the plane is shown in Fig. 5.1. If the initial velocity of the body is 1.5 m/s and =0.2, what velocity will have after moving 5 m?

Fig. 5.12. Two weight 50 N and 30 N are connected by thread and move along a rough horizontal plane under the action of a force 40 N applied to the first weight of 50 N is shown in Fig. 5.2. The coefficient of friction between the weight and plane is 0.4. Determine the velocity of the system after 3 seconds. Also calculate the tension in the string using Impulse-momentum equation.

Fig. 5.2

3. (i) Derive an equation for impulse momentum equations? (8 Marks)(ii) Derive an equation for work energy principle? (8 Marks)4. Two steel blocks, shown in Fig. 5.3, slide without friction on a horizontal surface. The velocities of the blocks immediately before impact are as shown. If the coefficient of restitution between the blocks is 0.75, determine (i) the velocities of the blocks after impact and (ii) the energy loss during impact?

Fig. 5.35. Two blocks A and B of masses MA= 280 kg and MB = 420 kg are joined by an inextensible cable is shown in Fig. 5.4. Assume that the pulley is frictionless and = 0.3 between block A and the surface. The system is initially at rest. Determine a) acceleration of block A b) velocity after it has moved 3.5 m and c) velocity after 1.5 seconds?

Fig. 5.4

6. A stone is projected with a speed of 30 m/s at an angle of elevation of 50. Find its velocity (i) After two seconds (ii) At the highest point of its path(iii) At a height of 6 m Find also the time interval between the two points at which the stone attains a speed of 23 m/s.?7. Two rough planes inclined at 30 and 60 to the horizontal and the same height are placed back to back. Masses of 12 kg and 24 kg are placed on the faces and are connected by a string passing over the pulley on the top of planes as shown in Fig. 5.5. if =0.6, find the resulting acceleration. MA = 12 kg, MB = 24 kg?

Fig. 5.58. Two Blocks A and B of weight 100 N and 200 N respectively are initially at rest on a 30 inclined plane as shown in Fig. 5.6. The distance between the blocks is 6 m. The co efficient of friction between the block A and the plane is 0.25 and that between the block B and the plane is 0.15. If they are released at the same time, in what time the upper block (B) reaches the Block (A)?

Fig. 5.6