1D  motion

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1. Introduction

In our dialy life ,we see lots

of things moving around for example car passing through from one place to

other ,person riding on a bicycle and many more like this.

In scientific terms an object

is said to be in motion ,if it changes its position with the pasage of

time and if it does not change it position with the passage of time then

it is said to be at rest

Both the motion and rest are

relative terms for example mobile kept on the table is resting at its

position but it is moving in the sense as earth is rotating on its axis.So

for a person seeing mobile from earth it is at rest and for person on moon

earth seems to change its position with time and so mobile is moving.

Simplest case of motion is

rectilinear motion which is the motion of the object in a straight line

In our descriotion of object

,we will treat the object as an point object

Object under consideration can

be treated as point object if the size of the object is much smaller than

the distance travelled by it in a reasonable time duration for example

length of a motor

car travelling a distance of 500km can be neglected w.r.t

distance travelled by it.

Here in kinematics ,we study

ways to describe the motion without going into the cause of the motion

2.Position and Displacement

(a) Position:

To locate the position in

motion or at rest,we need a frame of refrence.

Simplest way to choose a frame

of refrence is to choose three mutually perpendicular axis labelled as

X-,Y- and Z- axis as shown in figure below

Such system of labelling

position of an object is known as rectangular coordinates system

If A(x,y,z) be the position of

any point in rectangular co-ordinates system it can be labelled as follows

Point O is the point of

intersection of these mutaully perpendicular axis and is known as refrence

point or origin of frame of refrence

To measure a time ,we can also

attach a clock with this frame of refrence

If any or all co-ordinates of

the object under consideration changes with time in this frame of refrence

then the object is said to be in a motion w.r.t the frame of the refrence

otherwise it is at rest

For describing motion in one

dimension we need one set of co-ordinates axis i.e only one of X,Y and Z

axis

Similary for two and three

dimensions we need two or three set of axis respectively

Motion of an object along a straight line

is an example of motion in one dimension

For such a motion,any one axis

say X-axis may be choosen so as to co-incide with the path along which

object is moving

Position of the object can be

measured w.r.t origin O shown in the figure

Position to the right of the

origin has positive values and those to the left of origin O has negative

values.

(b) Distance and displacement:

In the graph shown below an

object is at position P at time t1 and at position R at

time t2.

In the time interval from t1 to

t2 particle has travelled path PQR and length of the path

PQR is the distance travelled by the object in the time interval t1 to

t2

Now connect the initial

position of the object P with its final position R through a straight line

and we get the displacement of the object.

Displacement of the object has

both magnitude and direction i.e., displacement is a vector quantity.

Magnitude of displacement

vector is equal to the length of straight line joining initial and final

position and its direction points from the initial position of object

towards its final position.

In contrast to displacement

distance is scalar quantity.

NEWTON’S LAWS OF MOTION Newton’ 1st law or Law of Inertia Every body continues to be in its state of rest or of uniform motion until and unless and until it is compelled by an external force to change its state of rest or of uniform motion. Inertia The property by virtue of which a body opposes any change in its state of rest or of uniform motion is known as inertia. Greater the mass of the body greater is the inertia. That is mass is the measure of the inertia of the body. Numerical Application If, F = 0 ; u = constant Physical Application 1. When a moving bus suddenly stops, passenger’s head gets jerked in the forward direction. 2. When a stationery bus suddenly starts moving passenger’s head gets jerked in the backward direction. 3. On hitting used mattress by a stick, dust particles come out of it. 4. In order to catch a moving bus safely we must run forward in the direction of motion of bus. 5. Whenever it is required to jump off a moving bus, we must always run for a short distance after jumping on road to prevent us from falling in the forward direction. Key Concept In the absence of external applied force velocity of body remains unchanged. Newton’ 2 nd law Rate of change of momentum is directly proportional to the applied force and this change always takes place in the direction of the applied force. dp F dt K.V. Silchar 29 or, dp =F (here proportionality constant is 1) dt putting, p = mv F = dp dt or, F = dmv dt or, F = mdv + vdm dt dt or, F = mdv (if m is constant dm/dt = 0) dt or, F = ma Note :- Above result is not Newton’s second law rather it is the conditional result obtained from it, under the condition when m = constant. Numerical Application a = FNet m Where FNet is the vector resultant of all the forces acting on the body. F1 F2 F6 m F3 m FNet F5 F4 Where, FNet = F1 + F2 + F3 + F4 + F5 + F6 K.V. Silchar 30 Physical Application Horizontal Plane i) Case - 1 N Body kept on horizontal plane is at rest. For vertical direction N = mg(since body is at rest) mg ii) Body kept on horizontal plane is accelerating horizontally under single horizontal force. N For vertical direction N = mg (since body is at rest) F For horizontal direction F = ma mg iii) Body kept on horizontal plane is accelerating horizontally towards right under two horizontal forces. (F1 > F2) N For vertical direction N = mg (since body is at rest) F2 F1 For horizontal direction F1 - F2 = ma mg iv) Body kept on horizontal plane is accelerating horizontally under single inclined force FSinθ F N For vertical direction N + FSinθ = mg (since body is at rest) θ FCosθ For horizontal direction FCosθ = ma mg v) Body kept on horizontal plane is accelerating horizontally towards right under an inclined force and a horizontal force. F1Sinθ a N F1 For vertical direction N + F1Sinθ = mg (since body is at rest) F2 θ F1Cosθ For horizontal direction F1Cosθ – F2 = ma mg a a a a K.V. Silchar 31 vi) Body kept on horizontal plane is accelerating horizontally towards right under two inclined forces acting on opposite sides. N F1Sinθ F1 For vertical direction N + F1Sinθ = mg + F2 SinФ (since body is at rest) F2CosФ Ф θ For horizontal direction F1Cosθ F1Cosθ – F2CosФ = ma F2 F2SinФ mg Inclined Plane i) Case - 1 N Body sliding freely on inclined plane. Perpendicular to the plane N = mgCosθ (since body is at rest) mgSinθ θ Parallel to the plane mgCos θ mgSinθ = ma θ mg ii) Case - 2 Body pulled parallel to the inclined plane. N F Perpendicular to the plane N = mgCosθ (since body is at rest) mgSinθ Parallel to the plane θ F - mgSinθ = ma mgCos θ mg θ iii) Case - 3 Body pulled parallel to the inclined plane but accelerating downwards. N Perpendicular to the plane F N = mgCosθ (since body is at rest) Parallel to the plane mgSinθ θ mgSinθ - F = ma mgCos θ θ mg a a a a K.V. Silchar 32 iv) Case - 4 Body accelerating up the incline under the effect of two forces acting parallel to the incline. N F1 Perpendicular to the plane N = mgCosθ (since body is at rest) F2 Parallel to the plane mgSinθ θ F1 - F2 - mgSinθ = ma mgCos θ mg θ v) Case - 5 Body accelerating up the incline under the effect of horizontal force. F1Cos θ N θ F1 Perpendicular to the plane N = mgCosθ + F1Sinθ (since body is at rest) F1Sin θ Parallel to the plane mgSinθ F1Cosθ - mgSinθ = ma mgCos θ mg θ vi) Case - 6 Body accelerating down the incline under the effect of horizontal force and gravity. N FSinθ Perpendicular to the plane N + FSinθ = mgCosθ (since body is at rest) F θ FCosθ Parallel to the plane mgSinθ FCosθ + mgSinθ = ma mgCos θ mg a θ vii) Case - 7 Body accelerating up the incline under the effect of two horizontal forces acting on opposite sides of a body and gravity. N F2Cosθ F1Sinθ θ F2 Perpendicular to the plane F1 N + F1Sinθ = mgCosθ + F2Sinθ(since body is at rest) θ F2Sinθ F1Cosθ Parallel to the plane mgSinθ F2Cosθ - F1Cosθ - mgSinθ = ma mgCos θ mg θ mg a a a a K.V. Silchar 33 Vertical Plane i) Case - 1 Body pushed against the vertical plane by horizontal force and moving vertically downward. For horizontal direction mg = ma (since body is at rest) N F For vertical direction F = N mg ii) Case - 2 Body pushed against the vertical plane by horizontal force and pulled vertically upward. F2 For vertical direction F2 - mg = ma N For horizontal direction (since body is at rest) F1 N = F1 mg iii) Case - 3 Body pushed against the vertical plane by inclined force and accelerates vertically upward. FCos θ F For horizontal direction θ N = FSinθ (since body is at rest) FSinθ N For vertical direction FCosθ – mg = ma mg iv) Case - 3 Body pushed against the vertical plane by inclined force and accelerates vertically downward. N For horizontal direction FSinθ N = FSinθ (since body is at rest) θ F For vertical direction FCosθ FCosθ + mg = ma mg a a a a K.V. Silchar 34 Tension In A Light String Force applied by any linear object such as string, rope, chain, rod etc. is known as it’s tension. Since string is a highly flexible object so it can only pull the object and can never push. Hence tension of the string always acts away from the body to which it is attached irrespective of the direction. Tension of the string, being of pulling nature, always acts away from the body to which it is attached Physical Application i) Flexible wire holding the lamp pulls the lamp in upward direction and pulls the point of suspension in the downward direction. ii) Rope holding the bucket in the well pulls the bucket in the upward direction and the pulley in the downward direction. iii) Rope attached between the cattle and the peg pulls the cattle towards the peg and peg towards the cattle. iv) When a block is pulled by the chain, the chain pulls the block in forward direction and the person holding the chain in reverse direction. Key Point In case of light string, rope, chain, rod etc. tension is same all along their lengths. T1 P T2 Consider a point P on a light (massless) string. Let tensions on either side of it be T1 and T2 respectively and the string be accelerating towards left under these forces. Then for point P T1 - T2 = ma Since string is considered to be light mass m of point P is zero or, T1 - T2 = 0 or, T1 = T2 K.V. Silchar 35 i) Case - 1 Two bodies connected by a string are placed on a smooth horizontal plane and pulled by a horizontal force. N2 N1 m2 T T m1 F m2g m1g For vertical equilibrium of m1 and m2 N1 = m1g and N2 = m2g For horizontal acceleration of m1 and m2 F – T = m1a and T = m2a (Since both the bodies are connected to the same single string they have same acceleration) ii) Case - 2 Two bodies connected by a horizontal string are placed on a smooth horizontal plane and pulled by a inclined force. N2 N1 FSinθ F m2 T T m1 θ FCosθ m2g m1g For vertical equilibrium of m1 and m2 N1 + FSinθ = m1g and N2 = m2g For horizontal acceleration of m1 and m2 FCosθ – T = m1a and T = m2a (since both the bodies are connected to the same single string they have same accelerations) iii) Case - 3 Two bodies connected by a inclined string are placed on a smooth horizontal plane and pulled by a inclined force. a a K.V. Silchar 36 N2 N1 FSinθ F TCosθ θ m2 TSinθ T m1 θ T TSinθ FCosθ θ TCosθ m2g m1g For vertical equilibrium of m1 and m2 N1 + FSinθ = m1g + TSinθ and N2 + TSinθ = m2g For horizontal acceleration of m1 and m2 FCosθ – TCosθ = m1a and TCosθ = m2a (since both the bodies are connected to the same single string they have same accelerations) iv) Case - 4 Two bodies connected by a string made to accelerate up the incline by applying force parallel to the incline. N1 F m1gSinθ N2 T T m1gCosθ m1g m2gSinθ m2g m2gCosθ θ For equilibrium of m1 and m2 in the direction perpendicular to the plane N1 = m1gCosθ and N2 = m2gCosθ For acceleration of m1 and m2 up the incline F - T - m1gSinθ = m1a and T - m2gSinθ = m2a Tension of A light Rigid Rod Force applied by rod is also known as its tension. Since rod is rigid, it cannot bend like string. Hence rod can pull as well as push. Tension of rod can be of pulling as well as pushing nature but one at a time. Tension of a rod attached to the body may be directed towards as well as away from the body. T T FFF T T Tension of rod is pulling both the blocks Tension of rod is pushing both the blocks a a K.V. Silchar 37 Physical Application i) Pillars supporting the house pushes the house in the upward direction and pushes the ground in the downward direction. ii) Wooden bars used in the chair pushes the ground in the downward direction and pushes the seating top in the upward direction. iii) Parallel bars attached to the ice-cream trolley pushes the trolley in the forward direction and pushes the ice-cream vendor in the backward direction.(when the trolley is being pushed by the vendor) iv) Rod holding the ceiling fan pulls the fan in the upward direction and pulls the hook attached to the ceiling in the downward direction. v) Parallel rods attached between the cart and the bull pulls the cart in the forward direction and pulls the bull in the backward direction. Different Cases of Light Rigid Rod i) Case - 1 Rod attached from the ceiling and supporting the block attached to its lower end. Since the block is at rest T T = mg T m mg ii) Case - 2 Rod is attached between two blocks placed on the horizontal plane and the blocks are accelerated by pushing force. N1 N2 For vertical equilibrium of m1 and m2 m1 T T m2 N1 = m1g and N2 = m2g F For horizontal acceleration of m1 and m2 F – T = m1a and T = m2a m1g m2g (Since both the bodies connected to the rod will have same acceleration) iii) Case - 3 Rod is attached between two blocks placed on the horizontal plane and the blocks are accelerated by pulling force. N2 N1 m2 T T m1 F For vertical equilibrium of m1 and m2 N1 = m1g and N2 = m2g m2g m1g a a K.V. Silchar 38 For horizontal acceleration of m1 and m2 F – T = m1a and T = m2a (Since both the bodies are connected to the same rod they have same acceleration) iv) Case - 4 Rod is attached between two blocks placed on the incline plane and the blocks are accelerated by pushing parallel to the incline. N2 m2gSinθ For vertical equilibrium of m1 and m2 N1 T N1 = m1gCosθ and N2 = m2gCosθ T m2gCosθ m2ga For acceleration of m1 and m2 parallel to F m1gSinθ the incline m1gCosθ F – m1gSinθ - T = m1a, θ m1g T – m2gSinθ = m2a Fixed Pulley It is a simple machine in the form of a circular disc or rim supported by spokes having groove at its periphery. It is free to rotate about an axis passing through its center and perpendicular to its plane. Key Point In case of light pulley, tension in the rope on both the sides of the pulley is same (to be proved in the rotational mechanics) r r T1 T2 Anticlockwise Torque - Clockwise Torque = Moment of Inertia x Angular acceleration T1 x r - T2 x r = Iα Since the pulley is light and hence considered to be massless, it’s moment of inertia I = 0 or, T1 x r - T2 x r = 0 or, T1 x r = T2 x r or, T1 = T2 a K.V. Silchar 39 Different Cases of Fixed Pulley i) Case - 1 Two bodies of different masses (m1 > m2) are attached at T1 two ends of a light string passing over a smooth light pulley For vertical equilibrium of pulley T1 T1 = T + T = 2T T T For vertical acceleration of m1 and m2 m1g - T = m1a and T - m2g = m2a T T m1 accelerates downwards and m2 accelerates upwards(m1>m2) m1 m2 m1g m2g ii) Case - 2 Two bodies of different masses are attached at two ends of a light string passing over a light pulley. m1 is placed on a horizontal surface and m2 is hanging freely in air. N For vertical equilibrium m1 N = m1g m1 T T For horizontal acceleration of m1 T = m1a m1g T For vertically downward acceleration of m2 T m2g - T = m2a m2g iii) Case - 3 Two bodies of different masses are attached at two ends of a light string passing over a light pulley. m1 is placed on an inclined surface and m2 is hanging freely in air. a a a a K.V. Silchar 40 For equilibrium of m1 perpendicular to incline plane T N = m1gCosθ N T For acceleration of m1 up the incline plane T T - m1gSinθ = m1a m1 m1gSinθ m2 For vertically downward acceleration of m2 m2g - T = m2a m2g m1g m1gCosθ θ Movable Pulley The pulley which moves in itself is known as movable pulley. Key Point In case of light movable pulley, acceleration of a body (pulley) goes on decreasing on increasing the number of strings attached to it. That is the body attached with two ropes moves with half the acceleration of the body attached with single rope. Length of the string is constant z x + 2y + z = L (Constant) Differentiating both sides with respect to t (Time) dx + 2dy + dz = dL dt dt dt dt y or, v1 + 2v2 + 0 = 0 (z and L are constant) x or, v1 + 2v2 = 0 Again differentiating both sides with respect to t dv1 + 2dv2 = 0 dt dt m1 m2 or, a1 + 2a2 = 0 or, a1 = - 2a2 That is acceleration of m1 (body attached to a single string) is opposite and twice the acceleration of m2 (body attached to a double string) Different Cases of Light Movable Pulley i) Case - 1 Mass m1 is attached at one end of the string and the other end is fixed to a rigid support. Mass m2 is attached to the light movable pulley. a1 a2 a a K.V. Silchar 41 w T1 T For vertical acceleration of m1 T1 m1g - T = m12a (m1 is connected to a single string) For vertical acceleration of m2 T T T2 – m2g = m2a (m1 accelerates downwards and m2 accelerates upwards since m1>2m2) T T T2 For the clamp holding the first pulley T1 = 2T T T2a For the clamp holding the movable pulley m1 m2 2T - T2 = mpulleya or, 2T - T2 = 0 (light pulley) or, 2T = T2 m1g m2g ii) Case - 2 Mass m1 is attached at one end of the string and placed on a smooth horizontal surface and the other end is fixed to a rigid support after passing through a light movable suspended pulley. Mass m2 is attached to the light movable pulley. N For vertical equilibrium of m1 N = m1g m1 T T T For horizontal acceleration of m1 T = m12a m1g T For vertical motion of m2 T T m2g – 2T = m2a m2 m2g iii) Case - 3 Mass m1 is attached to the movable pulley and placed on a smooth horizontal surface. One end of the string is attached to the clamp holding the pulley fixed to the horizontal surface and from its other end mass m2 suspended. N For vertical equilibrium of m1 T T N = m1g m1 T T For horizontal motion of m1 T 2T = m1a m1g T For vertical motion of m2 m2 m2g - T = m22a m2g a a a 2a 2a a 2a a K.V. Silchar 42 iv) Case - 4 Mass m1 is attached to a movable pulley and placed on a smooth inclined surface. Mass m2 is is suspended freely from a fixed light pulley. t T T For equilibrium of m1 perpendicular to incline plane T N = m1gCosθ T x T T For acceleration of m1 up the incline plane N T m2 2T - m1gSinθ = m1a m1 For vertically downward acceleration of m2 m2g m2g - T = m22a m1gSinθ m1gCosθ m1g θ Newton’ 3 rd law or Law of Action and Reaction Every action is opposed by an equal and opposite reaction. or For every action there is an equal and opposite reaction. F12 m1 F21 m2 F12 is the force on the first body (m1) due to second body (m2) F21 is the force on the second body (m2) due to first body (m1) If F12 is action then F21 reaction and if F21 is action then F12 reaction Numerical Application Force on the first body due to second body (F12) is equal and opposite to the force on the second body due to first body (F21). F21 = - F12 Physical Application i) When we push any block in the forward direction then block pushes us in the backward direction with an equal and opposite force. ii) Horse pulls the rod attached to the cart in the forward direction and the tension of the rod pulls the cart in the backward direction. a 2a K.V. Silchar 43 iii) Earth pulls the body on its surface in vertically downward direction and the body pulls the earth with the same force in vertically upward direction. iv) While walking we push the ground in the backward direction using static frictional force and the ground pushes us in the forward direction using static frictional force. v) When a person sitting on the horse whips the horse and horse suddenly

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