Linear Motion, Newton's Laws, Friction

Linear motion 

When an object moves along a line, it is said to undergo linear motion. If it moves along a circular path or a bend, it is said to undergo circular or rotational motion. If the object moves to-and-fro about a mean position, it is said to undergo vibratory or oscillatory motion. 

There are various terms associated with linear motion: 

Displacement (s): This refers to the distance covered by an object in a given direction. Displacement is a vector quantity hence has both direction and magnitude (distance has magnitude only). For example, suppose an object starts at point A, moves to point B, a distance of 10 m left of point A, then to point C, a distance of 20 cm right of B.

To obtain the total displacement, the direction has to be put into consideration; If the displacement to the left is considered as negative, the opposite direction (right) is taken as positive. The total displacement (s) becomes;

The total displacement is 10 cm to the right of A (since answer is positive).

To evaluate the distance. the direction is disregarded hence total distance is given as:

Velocity (v): This refers to the rate of change of displacement with time. It is a vector quantity with the same direction as the displacement. A body is said to move at a constant velocity if it covers equal distances within equal intervals of time. A graph of displacement against time in this case is a straight line inclined to the horizontal.

Suppose the displacement of a body changes uniformly by ∆s  in time ∆t . The velocity v of the body becomes:

                                                         (i)

If a body is stationary (not moving), the graph of displacement against time is a straight horizontal line.

Acceleration (a): Suppose displacement of a body is not uniform over equal intervals of time. 

In this case, the body is said to be accelerating. Acceleration is therefore the rate of change of velocity with time. If the change in velocity is equal in equal intervals of time, the acceleration is said to be uniform and a graph of velocity against time is a straight line inclined to the horizontal.

If a body initially moving with a velocity u (initial velocity) accelerates uniformly to a velocity v (final velocity) in time t, the acceleration a of the body is given by; 

                                           (ii)

If the body is not accelerating (i.e., body moving at a constant velocity, velocity not changing with time) the velocity-time graph is a horizontal line.

If the velocity of the body reduces with time, for example when a vehicle slows down to a stop, it is said to be decelerating or with a negative acceleration given that the initial velocity is greater than the final velocity. A graph of velocity against time in this case is a straight lined inclined to the vertical. 

Equations of linear motion

The relationship between velocity, displacement and acceleration of a body undergoing linear motion are represented by three main equations referred to as equations of linear motion. Consider a car initially moving with a velocity u. Suppose too that in time t, the velocity of the car increases uniformly from u to v during which time it undergoes a displacement s

It follows that the acceleration a is equal to; 

      

                                (iii)

Equation (iii) is the first equation of linear motion.

Now, the average velocity vav of the car is equal to;

                                          (iv)

The displacement s of the car in time t is the product of average velocity and time, i.e.,

                                        (v)

Now, equation (iii) may be expressed as;

                                                (vi)

Using equation (vi) in equation (v) leads to;

    (vii)

                                           (viii)

                                (ix)

Equation (ix) represents the second equation of linear motion.

Using equation (iii) in equation (v) leads to;

        (x)

                               (xi)

                         (xii)

Equation (xii) represents the third equation of linear motion.

In a nutshell, the three equations of linear motion are;

NOTE

(i) The area under a velocity-time graph is equal to the total displacement of a body moving with constant acceleration, and equal to the total distance if the acceleration changes with time.

(ii) For objects under free fall (influence of gravity only) for example when a stone is thrown up or dropped from a higher elevation, the acceleration due to gravity (g) is used. The value of g is negative when the body is moving upwards (decelerating) and positive when the body is moving downwards (accelerating). 

Motion under gravity

Suppose a body is launched straight up with a velocity u. Say the body reaches a maximum height s in time t. At maximum height therefore, the final velocity v equals zero. It follows that by the first equation of linear motion:

                                       (i)

                                        (ii)

                                            (iii)

Equation (iii) represents time taken by the body to reach maximum height.

Since the velocity at maximum height is zero, then by the second equation of linear motion;

                                  (iv)

                                          (v)

                                                (vi)

Equation (vi) represents the maximum height reached by the body.

If the body goes up then back to its launching point, the total displacement is zero and therefore by the third equation of linear motion it follows that; 

                               (vii)

                                       (viii)

                                           (ix)

                                               (xi)

Equation (xi) represents time of flight. The equation shows that the time of flight is twice the time taken to reach the maximum height. 

Energy of  object under the influence of gravity

An object moving under the action of gravity posses both potential energy, PE (by virtual of its distance from the earth's surface) and kinetic energy, KE (by virtual of its motion). 

Potential energy of an object is a product of its weight and the distance from the earth's surface. Weight, W, of an object is the product of the mass of the object, m, and acceleration due to gravity, g, i.e.

                                         (i)

If an object is at a  height h above the earth's surface, its potential energy PE is given;

                                     (ii)

Kinetic energy of a body is a functions of its motion and increases with the velocity. For a body of mass m moving with a constant velocity v, the kinetic energy KE is given by;;

                                              (iii)

A stone moving up in the air posses both potential energy and kinetic energy. The sum of potential energy and kinetic energy is called mechanical energy (ME), that is;

                                      (iv)

Consider a ball projected vertically upwards  with a velocity v.  If the effect of air-resistance is ignored, the mechanical energy of the ball as it moves up and down remains constant. This means that when KE=maximum, PE=0. This occurs just after the ball is launched. At this point in time, the ball has maximum velocity (v is maximum) but least distance from the ground (h is zero). Similarly, when KE=0, PE=Max. This occurs when the ball reaches maximum height (just before it starts moving downwards). At this point, the velocity of the ball is zero while the height above ground is maximum. The ball has both KE and PE anywhere else between launch and maximum height.

NOTE

Equations of motion represent the kinematics of motion, showcasing relationships between displacement, velocity, and acceleration. Objects at rest do not spontaneously move, neither do they spontaneously accelerate or decelerate if in motion. An external force must interact with the body for change of state of motion to occur. Newtons laws of motion looks at the dynamics of an object with respect to the forces acting on it and the subsequent consequences. The laws therefore describe the kinetics of an object.

Newton's laws of motion

There are three Newton’s laws of motion

Newton’s first law of motion 

Also referred to as the law of inertia (inertia is the tendency of a body to resist change). It states that; a body remains at rest or in motion with constant velocity along a straight line unless acted upon by an external force. For example, say a stone tied to a string is swirled around in a circular path with a linear velocity v (tangential to the circular path). The tension in the string keeps the stone in the circular path and if the string snaps, the stone flies straight away with a constant velocity. 

Newton’s second law of motion

This states that the rate of change of momentum is directly proportional to the force applied and takes place in the direction of the force. Consider a toy car, mass m, initially moving with a velocity u. Suppose too that a force F is applied on the toy car for a time during which time its velocity increases uniformly from u to v. 

Considering that momentum is the product of mass and velocity, the rate of change of momentum with time, ∆P/∆t, of the toy car is given as;

                                          (i)

By Newton’s second law;

                                                     (ii)

                                            (iii)

                                       (iv)

where k is a constant of proportionality.

                    (v)

Now, ∆v/∆t = a (acceleration) hence

                                              (vi)

k is assigned a value of one (unity) when force is in Newtons (N), mass in kg and acceleration in m/s2 hence:

                                                 (viii)

Equation (viii) is the mathematical representation of Newton’s second law of motion.

Newton’s third law of motion

This states that action and reaction are equal and opposite. A body exerts force on the surface it is resting on. Likewise, the surface exerts an equal but opposite force called reaction R on the body. If the surface is horizontal, the reaction is equal to the weight of the object. If on the other hand the surface is inclined, say at an angle θ to the horizontal, the reaction is equal to the component of weight that is perpendicular to the surface. For a mass m resting on a plane inclined at an angle θ therefore, the reaction R is given by;

                                     (ix)

Applications of action and reaction:  rowing a boat - when water is pushed backwards, the boat moves forward; balloon flies - air gushes out from the rear and balloon moves in the opposite direction.

NOTE

(i) The component of weight of the body along the plane inclined at an angle θ (mg sin⁡θ) is responsible for motion down the plane. 

If the body accelerates down the plane with an acceleration a, by Newton’s second law of motion (and assuming no frictional force exists), it follows that:

                                   (x)

Considering that sin⁡ θ = 0 (minimum) when θ = 0 and  sin ⁡θ = 1 (maximum) when θ = 900  it follows that the steeper the inclination, the higher the acceleration of the body.

Impulse and conservation of momentum

Equation (v) can be expresses as;

                                (xi)

If ∆t is very small (e.g., when a bat strikes a tennis ball) then;

                                  (xii)

                      (xiii)

By equation (xiii) impulse of a force is equal to the change in momentum of an object when a force is applied for a very short period. 
If the net force in equation (xi) is zero (F =0) then;

                                     (xiv)

                                             (xv)

It follows from equation (xv) that if no external force acts on a body, the final momentum is equal to the initial momentum (momentum is not changing with time) hence momentum is conserved.

Collisions

When two objects collide in the absence of external forces (such as friction), two things may happen:

(1) The colliding objects may separate after collision and move in such a way that both momentum and kinetic energy (KE) are conserved, i.e.

  • Momentum before collision = momentum after collision
  • KE before collision = KE after collision

Such a collision is called an elastic collision. Virtually no energy is lost during an elastic interaction.  

(2) The colliding particles may merge (or separate) and move in such a way that momentum is conserved but kinetic energy is not, i.e. 

  • Momentum before collision = momentum after collision
  • KE before collision ≠ KE after collision

Such a collision is referred to as an inelastic collision. During an inelastic collision, part of the initial kinetic energy is converted to heat, sound, deformation work etc. 

Mass and Weight

Mass is the quantity of matter contained in an object. The quantity of matter does not change with location. As an example, the mass of say a basket of ten balls will remain constant whether the mass is measured on the moon or on earth. Mass is also defined as a measure of inertia of an object (inertia is the tendency of an object to resist change) in accordance with Newton’s first law of motion (also referred to as law of inertia) which states that a body at rest remains at rest, and if in motion travels at a constant velocity in a straight line, unless an external force acts on it). An object with a large mass (heavy) has a high inertia compared to one with low mass (light). It is therefore easier to lift a light object compared to a heavy object. It is also easier to stop a lighter object from moving (or change direction) compared to a heavier one. Mass is a scaler quantity described in terms of quantity only and not direction.

Weight on the other hand is a vector quantity and as such has both magnitude and direction. It is a measure of the gravitational force of attraction exerted on an object by a massive body, for example the earth or the moon, and is usually directed towards the center of the attracting body. This means that if an object is placed at some height above the massive body, or on a slope of a smooth plane within the massive body, and if there is no frictional force acting on the object, then the object will accelerate uniformly towards the lowest level in accordance with Newton’s second law of motion. According to Newton’s second law of motion, the rate of change in momentum of an object is proportional to the force applied and takes place in the direction of the force. It therefore follows that if some force F acts on an object of mass m for a time and within this time the velocity of the object changes from u to v in the direction of the applied force, then:

Since

where a is the acceleration , it follows that;

Similarly, the gravitational force (weight, W) is the product of mass (m) of the object and acceleration due to gravity of the attracting body (g), i.e. 

The acceleration due to gravity is not constant. On the earth’s surface for instance, the acceleration due to gravity is lowest at the equator with a value of 9.798 m/s2 and increases with distance away from the equator reaching a maximum at the poles with a value of 9.862 m/s2. The variation in earth’s gravitational acceleration is due to the fact that the surface of the earth at the poles is closer to the center of the earth (which is considered to be the source of the gravitational force of attraction) than the surface of the earth at the equator. This is because the earth is not a perfect sphere but rather an oblong sphere that bulges around the equator. An object of mass 1 kg will thus weigh 9.798 N at the equator and 9.862 N at the poles. Nonetheless, an average value of 1.82 m/s2 is adopted as the earth’s gravitational acceleration. The gravitational acceleration on the moon is 1.62 m/s2 which is lower than the value on earth. A 1 kg object on the moon would therefore weigh 1.62 N.

Effective weight (force exerted on the floor) when travelling in a lift

(1) Lift stationary or moving with constant velocity, v: The force an object, say mass m, exerts on the floor of the lift is equal to its weight W, i.e.,

                                              (i)   

(2) Lift moving downwards with acceleration a; In this case, both the weight of the object (mg) and the external force on the object on account of an accelerating lift (ma) are in the same direction. The effective force W’ the object exerts on the floor of the lift therefore equals;

      (ii)

Equation (ii) shows that an object weighs less than it really is if weighed while in a lift travelling downwards, and if g>a. If the acceleration of the lift equals the acceleration due to gravity i.e., g=a, then;

                                                  (iii)

This means that the object does not exert any force on the floor of the lift and therefore appears weightless. If g<a, the speed of the lift changes at a higher rate than that of the object hence W’is negative;

                                                  (iv)   

The object in this case appears to move upwards relative to the lift. 

(3) Suppose that the lift is moving upwards with constant acceleration a. The force on the object by virtual of gravity (mg) and that due to the external force (ma) are in opposite directions. The effective force W’’ the object exerts downwards is therefore given by:

     (v)

The object in this case appears to weigh more than it actually is.

In a nutshell:

Friction

Friction is opposition to motion between two surfaces that are sliding or attempting to slide over each other. It acts in the direction opposite that of motion.

If frictional force equals the applied force, the body moves at a constant velocity (has zero acceleration). If applied force is greater than the frictional force, the body accelerates and if less than frictional force the body decelerates. 

Friction is of two types:

  1. Static/limiting friction
  2. Sliding/dynamic/kinetic friction

(1) Static friction

This is the opposition force between two surfaces that are just about to slide over each other. 

To demonstrate static friction, consider the diagram shown. Suppose that small weights are gradually added to the pan.

Initially mass A does not move. As more masses are added to the pan, a point is reached when the mass just starts to move. At this point, the  static frictional force (F)  is equal to the tension (T) in the string which is equal to the mass of the pan and its contents.

Alternatively, consider a mass m lying a horizontal plank of wood. The plank is then tilted gradually  until the mass just starts to slip. The angle, say θ, (called the angle of friction)  that this happens is noted.

The static frictional force, say F', in this case is given by;

Note: Coefficient of static friction µ’ is defined as the ratio of static friction F' to the normal reaction R that is;

(2) Sliding 

This is the frictional force between two surfaces in contact and in relative motion. Sliding friction is usually lower than static force. This implies that more force is required to start a body moving compared to that required to keep the body moving. The extra force is used to overcome inertia. 

Consider the diagram shown where F'' represents the sliding friction. 

If the body moves at a constant velocity, the force down the plank is equal to the sliding friction, i.e. 

The body therefore does not accelerate. 

If on the other hand;

the body experiences a net force down the plane equal to;

This makes the body accelerate downwards. If a be the acceleration, then;

Thus, the acceleration of an object down a tilted plane can be increased by either; 

  • increasing the angle of tilt (angle of friction)
  • reducing the frictional force for example by using a smother plank.

NOE: the coefficient of sliding friction µ’’ is given by;

Advantages of friction– allows us to walk (cannot walk on slippery floor), write, drive (if road smooth, the vehicle skids).

Disadvantages of friction -causes wear and tear in moveable parts

Ways of reducing friction – oiling, using rollers and ball bearings. 


EXAMPLES


Example 1 (KCSE 2021)

(a) A bus moving initially at a velocity of 20 m/s decelerates uniformly at 2 m/s2 

(i) Determine the time taken for the bus to come to a stop. (3 marks)

u = 20 m/s, v = 0 m/s

 

(ii) Sketch the velocity-time graph for the motion of the bus up to the time it stopped. (2 marks)

Answer

(iii) Use the graph to determine the distance moved by the bus before stopping. (1 mark)

Answer

(2) A car of mass 1000kg travelling at a constant velocity of 40 m/s collides with a stationary metal block of mass 800kg. This impact takes 3 seconds before the two move together. Determine the impulsive force. (4 marks)

Answer

First determine the final velocity, say v

momentum before collision = momentum after collision

The determine acceleration 

Then determine force:

(3)  State the reason why an object on earth has a higher weight than on the moon. (1 mark)

Answer: An object on earth has a higher weight than on the moon because the acceleration due to gravity on earth is higher than on the moon


Example 2 (KSCE 2020)

(a) Figure 6 shows the velocity-time graph of the motion of a stone thrown vertically upwards.

From the graph, determine the maximum height reached by the stone. (3 marks)

Answer

Initial velocity, u = 20 m/s

At maximum height, velocity (final velocity) v = 0 m/s

(b)  Figure 7 shows a box placed on a weighing balance. The balance is placed on the floor of a lift.

State what would be observed on the reading of the balance when the:

(i) lift is accelerating downwards (1 mark)

Answer: The effective weight is lower than the actual weight. The balance indicates a lower value

(ii) lift moves downwards with a uniform velocity (1 mark)

Answer: The effective weight is equal to the actual weight. Balance indicates the actual weight.

(iii) lift is accelerating upwards (1 mark)

Answer: The effective weight is greater than the actual weight. The balance reads a higher value.


Example 3 (KCSE 2019)

(a) On the axes provided, sketch a displacement — time graph for a trolley moving down a frictionless inclined plane till it reaches the end of the incline.   (1 mark)

Solution

Body is uniformly accelerating hence displacement-time graph curve upwards.

(b) State two ways in which an inclined plane can be made to reduce the applied effort when pulling a load along the plane. (2 marks)

Answer

  • The steeper the inclination, the higher the acceleration hence the higher the effort required to counter this motion. To reduce the effort therefore, the angle of inclination should be reduced.
  • Friction opposes motion and as such should be reduced for example by oiling the surface or placing the load on rollers.

(c)Two boxes E and F of masses 2.0kg and 4.0kg respectively are dragged along a frictionless surface using identical forces. State with a reason which box moves with a higher velocity acceleration. (2 marks)

Answer

From Newton's second law of motion

 

For box E;

  

For box F;

 

aE > aF 

E moves with a higher acceleration because it has a smaller mass.


Example 4 (KCSE 2018)

(a) State Newton’s first law of motion. (1 mark)

Answer: Law of inertia – a body remains at rest or in motion with constant velocity along a straight line unless acted upon by an external force.

(b) A wooden block resting on a horizontal bench is given an initial velocity u so that it slides on the bench for a distance x before it stops. Various values of x are measured for different values of the initial velocity. Figure 9 shows a graph of u2 against x.

(i) Determine the slope S of the graph. (3 marks)

Answer

Use (8, 50) and (0,0)

(ii) Determine the value of k given that u2 = 20kd where k is a frictional constant for the surface. (2 marks)

(iii) State with a reason what happens to the value of k when the roughness of the bench surface is reduced. (2 marks)

Answer: When roughness reduces, friction reduces hence the value of k reduces.

(c) An object is thrown vertically upwards with an initial velocity of 30 m/s. Determine its maximum height (acceleration due to gravity g is 10 m/s2). (3 marks)

Answer

(b)A stone is dropped from the top of a building to the ground. On the axes provided, sketch a graph of potential energy against time for the stone. (1 mark)

Answer

Gravitation potential energy PE is a function of heightabove the surface. The potential energy is maximum when height above the ground is maximum, reduces as h reduces and attains a minimum value when h is minimum. The graph curves as a result of acceleration due to gravity.


Example 5 (KCSE 2017)

A stone is thrown vertically upwards. Sketch a graph of potential energy (y axis) against time as the stone moves until it hits the ground. (1 mark)

Solution

(NOTE: Potential energy (PE) increases with distance above the earth’s surface, with PE on the earth’s surface (ground) taken to be zero. When a ball is thrown up in the air, the distance between the ball and the ground increases hence the PE energy increases as the ball moves up. The ball however slows down as it moves up due to the effect of gravity, coming to a momentarily stop at some maximum distance (height) from the earth’s surface. At this maximum height the ball has maximum PE. The ball then starts moving downwards under the influence of gravity, with the distance to the ground, and hence the PE, reducing with time. The PE drops to zero when the ball hits the ground).

(2) A tape attached to an accelerating trolley passes through a ticker timer that makes dots on it at a frequency of 50Hz. The ticker timer makes 10 dots on a 10cm long tape such that; the distance a between the first two dots is 0.5 cm and the distance b between the last two dots is 1.5 cm. 

(a) Determine the velocity of the trolley at:

(i) distance a, (4 marks)

Solution

 

 

But

Hence

         

(ii) distance b. (2 marks)

Solution

(ii) Determine the acceleration of the trolley                              (3 marks)

   

(b) state with reason what would be observed on the spacing between the dots if on the tape if the trolley is made to move on a horizontal surface.         (2 marks)

Answer

  • If frictional force was negligible, then the net force on the trolley would be zero hence the trolley would move at a constant velocity (zero acceleration). The dots would be equally spaced.
  • If on the other hand the frictional force was present, the speed of the trolley would reduce with time hence the spacing between the dots would reduce with time.

PRACTICE QUESTIONS


KCSE 2016

(1) The graph in Figure 10 shows the velocity of a car in the first 8 seconds as it accelerates from rest along a straight line.

Determine the distance travelled 3.0 seconds after the start. (2 marks)

(2) (c) A car of mass 800Kg starts from rest and accelerates at 1.2 m/s2. Determine its momentum after it has moved 400 m from the starting point. (3 marks)


KCSE 2015

(1) Mechanics is one of the branches of physics. State what it deals with (1 mark)

(2)  Figure 7 (drawn to scale) shows a section of tape after passing through a ticker timer operated at a frequency of 50 Hz. The tape is attached to a trolley moving in the direction shown.

(i) Determine the velocity between:

(I) P and Q; (4 marks)

(II) X and Y. (2 marks)

(ii) Determine the acceleration of the trolley. (3 marks)

(3) (b) Two bodies of masses 5 kg and 8 kg moving in the same direction with velocities 20 ms-1 and 15 ms-1 respectively collide in-elastically. Determine the velocity of the bodies after the collision. (4 marks)

(4)  A body is released from a height h. Sketch a graph of potential energy against kinetic energy as the body falls to the ground. (2 marks)


KCSE 2014

(1) State the reason why it is not correct to quote the weight of solid objects in kilograms. (1 mark)

(2) Figure 2 shows a section of a curved surface ABCD. Point A is higher than point B while BCD is horizontal. Part ABC is smooth while CD is rough. A mass m is released from rest at A and moves towards D.

State the changes in the velocity of m between:

(a) B and C; (1 mark)

(b) C and D. (1 mark)

(3) 15 Figure 9 shows a velocity-time graph for the motion of a body of mass 2 kg.

(a) Use the graph to determine the:(3 marks)

(i) displacement of the body after 8 seconds.(3 marks)

(ii) acceleration after point B;(3 marks)

(iii) force acting on the body in part (a) (ii).(2 marks)

(b) Sketch a displacement-time graph for the motion from point A to C.

(4) Figure 10 shows a trolley of weight 20 N pulled by a force of 4 N from the bottom to the top of an inclined plane at a uniform speed.

(ia State the value of the force acting downwards along the inclined plane.(1 mark)

(ib) Explain how the value in pan (a)  is obtained.(2 marks)


KCSE 2013

(1)A stone thrown vertically upwards reaches a height of 100 m. Determine the:

 (i) initial velocity of the stone. (2 marks)

(Neglect air resistance and take g = 10 ms-2)

(ii) total time the stone is in air. (2 marks)

(2) A horizontal force of 12 N is applied on a wooden block of mass 2 kg placed on a horizontal surface. It causes the block to accelerate at 5 ms-2. Determine the frictional force between the block and the surface. (3 marks)


KCSE 2012

(1) A student pulls a block of wood along a horizontal surface by applying a constant force. State the reason why the block moves at a constant velocity. (1 mark)

(2) A solid weighs 16.5 N on the surface of the moon. The force of gravity on the moon is 1.7Nkg-1. Determine the mass of the solid. (3 marks)

(3) Figure 3 shows a graph of velocity against time for a moving body.


Describe the motion of the body during the 10 seconds. (2 marks)

(4) Figure 9 shows a trolley on a smooth surface being pulled by a constant force F.

On the axis provided, sketch the velocity-time graph for the motion of the trolley. (2 marks)

(5) A ball of mass 200 g is thrown vertically upwards with velocity of 5 ms-1. The air resistance is 0.4 N.
Determine:

(a)The net force of the ball as it moves up; (take acceleration due to gravity g =10 ms-2) (2 marks)

(b) The acceleration of the ball. (3 marks)

(c) The maximum height reached by the ball. (3 marks)


KCSE 2011

(1) State the constant force that opposes the motion of a stone initially at rest, as it falls through air from a tall building. (1 mark)

(2) A particle starts from rest and accelerates uniformly in a straight line After 3 second it is 9m from the starting point. Determine the acceleration of the particle. (3 marks

(3) Figure 12 shows a lorry towing a trailer using a rope.

The lorry exerts a force N on the trailer and the trailer exerts an equal but opposite force M on the lorry. The frictional force between the trailer and the road is F. Explain how the forces N, M and F enable the trailer to move. (2 marks)

(4) Figure 13 shows a friction less trolley of mass 2kg moving With uniform velocity towards a wall. At the front of the trolley is a spring whose spring constant is 25N/m . The trolley comes to rest momentarily after compressing the spring by 3cm and then rebounds from the wall

Determine

(i) the force exerted on the Wall by the spring. (3 marks)

(ii) the maximum acceleration of the trolley as it rebounds from the wall. (3 marks)

)iii( State the reason why the trolley acquires a constant velocity alter it rebounds (2 marks)

(5) A bullet of mass 60g travelling at 800ms" hits a tree and penetrates a depth of 15 cm before coming to rest.

(i) Explain how the energy of the bullet changes as it penetrates the tree. (1 mark)

(ii) determine the average retarding force on the bullet. (3 marks)


KCSE 2010

)1) A stopwatch started 0.50s after the start button was pressed. The time recorded using the stopwatch for a ball bearing falling through a liquid was 2.53s. Determine the time of fall.    (2 marks)

(2) A cart of mass 30kg is pushed along a horizontal path by a horizontal force of 8N and moves with a constant velocity. The force is then increased to 14N. Determine:

     (a) The resistance to the motion of the cart.       (1 mark)

     (b) The acceleration of the cart.         (2 marks)

(3) A cyclist initially at rest moved down a hill without pedalling. He applied brakes and eventually stopped. State the energy changes as the cyclist moved down the hill.         (1 mark)

 (4) Figure 7 shows a mass of 30kg being pulled from point P to point Q, with a force of 200N parallel to an inclined plane. The distance between P and Q is 22.5m. In being moved from P to Q the mass is raised through a vertical height of 7.5m.

Determine the work done:

     (I) by the force;          (2 mark)

     (II) On the mass;        (2 marks)

     (III) To overcome friction (2 mark (ii) Determine the efficiency of the inclined plane.            (2 marks)

 (c) Suggest one method of improving the efficiency of an inclined plane.   (1 mark)

(5) Figure 9 shows a velocity-time graph for the motion of a certain body

Describe the motion of the region:

(i) OA;                  (1 mark)

(ii) AB;                  (1 mark)

(iii)BC.                  (1 mark)

(6) A car moving initially at 10m/s deceleration at 2.5m/s

(ia Determine:

(i) its velocity after 1.5s;                          (2 marks)

(ii)  the distance travelled in 1.5s;            (1 marks)

(iii) the time taken for the car to stop.    (2 marks)

(b) Sketch the velocity-time graph for the motion of the car up to the time the car stopped.          (1 mark)

c) From the graph, determine the distance the car travelled before stopping.    (2 marks)


 


Dr. Margaret W. Chege

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