Here students can locate TS Inter 1st Year Physics Notes 4th Lesson Motion in a Plane to prepare for their exam.
TS Inter 1st Year Physics Notes 4th Lesson Motion in a Plane
→ Vector: A physical quantity which has both magnitude and direction is called as vector. Ex: Displacement, Velocity, Force, etc.
→ Scalar: A physical quantity which has only magnitude is called as scalar.
Ex: Distance, Speed, Work, etc.
→ Equality of vectors: If two vectors are equal both in magnitude and direction are called ns-equal vectors.
→ Resultant vector : If the effect of many vectors is represented by a single vector then that single vector is called resultant vector.
→ Triangle law: If the magnitude and direction of two vectors are represented by two sides of a triangle taken in order then the third side of the triangle taken in reverse order will give the resultant both in magnitude and direction.
From the above figure, R = P + Q represents vectorial addition of P and Q.
→ Parallelogram law : If two vectors are represented by the two adjacent sides of a parallelogram then the diagonal passing through the intersection of those two vectors will represent the resultant both in direction and magnitude.
In the above figure, diagonal OB is sum of vectors i.e. R = A + B, diagonal AC is subtraction of vectors i.e. R = A – B
→ Laws of vector addition :
- Vector addition is commutative i.e.,
A̅ + B̅ = B̅ + A̅ - Vector addition obeys associative law i.e.,
(A̅ + B̅) + C̅ = A̅ + (B̅ + C̅)
→ Unit Vector: If the magnitude of any vector is unity then it is called unit vector.
Ex: unit vector A̅ = \(\frac{\overline{\mathrm{A}}}{|\overline{\mathrm{A}}|}\) = 1
Note: Unit vectors along X and Y directions are represented by i and j. In space unit vectors along X, Y and Z directions are represented by i, j and k.
→ Null vector: If the magnitude of a vector is zero then it is called null vector.
Null vector has only direction.
Ex: A̅ – A̅ = 0 If has only direction, magni¬tude is zero.
A̅ × 0̅ = 0̅ It has only direction, magnitude is zero.
→ Position vector: Any vector in a plane can be represented as A̅ = Ax i̅ + Ay j̅
Any vector in space can be represented as A̅ = Axi̅ + Ayj̅ + Azk̅
Where Ax, Ay and Az are magnitudes along X, Y and Z directions.
→ Resolution of vectors : Every vector can be resolved into two mutually perpendicular components. This division is with funda-mental principles of trigonometry.
A̅ = A̅x + A̅y
A̅ = A̅xî + A̅y ĵ
Ex Let A̅ makes an angle ‘O’ with X – axis then
X – component of \(\overline{\mathrm{A}}_{\mathrm{x}}=\overline{\mathrm{OB}}\) = A̅ cosO
Y-component of \(\overline{\mathrm{A}}_{\mathrm{y}}=\overline{\mathrm{OC}}\)=XsinO
Note: If values of A, and A are given then
Resultant A̅ = \(\sqrt{\mathrm{A}_{\mathrm{X}}^2+\mathrm{A}_{\mathrm{Y}}^2}\)
Angle made by vector A̅ with X-axis
θ = tan-1\(\left[\frac{A_Y}{A_X}\right]\)
→ Projectile : When a body is thrown into the space with some angle 0 (θ ≠ 90) to the horizontal it moves under the influence of gravity then it is known as projectile.
Note: The path of a projectile can be represented by the equation
y = ax – bx2. It represents a parabola.
Time taken to reach maximum height
t = \(\frac{v_0 \sin \theta}{g}\)
Maximum height reached hmax = \(\frac{v_0^2 \sin ^2 \theta}{2 g}\)
→ Time of flight (T) : The time interval from the instant of projection to the instant where it crosses the same plane or it touches the ground is defined as time of flight.
Time of flight T = \(\frac{v_0 \sin \theta}{g}\)
Note: For horizontally projected projectiles
T = \(\frac{v_0 \sin \theta}{g}\)
→ Range (or) horizontal range (R) : It is the horizontal distance from the point of projection to the point where it touches the ground.
Range R = \(\frac{2 \mathrm{v}_0^2 \operatorname{Sin} 2 \theta}{\mathrm{g}}\)
For horizontal projection R = v0\(\sqrt{2 h / g}\)
→ Uniform circular motion : If a body moves with a constant speed on the periphery of a circle then it is called uniform circular motion.
→ Time period: In circular motion time taken to complete one rotation is defined as time period (T).
Time period (T) = 2π/ω
Note: Frequency υ = \(\frac{1}{T}\) is equal to number of rotations completed in one second. Relation between ω and υ is ω = 2πυ or
v = 2πυ R.
→ Relative velocity in two-dimensional motion: Let two bodies A and B are moving with velocities V̅A and V̅B then relative velocity of A w.r.t B is V̅AB = V̅A – V̅B
Relative velocity of B w.r.t. A is
V̅BA = V̅B – V̅A
→ For (like) parallel vectors say P̅ and Q̅ resultant R̅ = P̅ + Q̅
→ For antiparallel vectors say P̅ and Q̅ resultant R̅ = P̅ – Q̅
→ Rectangular components of a vector R̅ are Rx = R cos θ and Ry = R sin θ
→ Resultant of vectors is given by parallelogram law.
(a) Resultant, (R) = \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}\)
(b) Angle made by Resultant
α = tan-1\(\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta}\)
(c) Difference of vectors
= \(\sqrt{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{PQ} \cos \theta}\)
where θ is the angle between P̅ and Q̅.
→ If two vectors a̅ and b̅ are an ordered pair then from triangle law, resultant R̅ = a̅ + b̅
(a) When two bodies A, B are travelling in the same direction ⇒ relative velocity, VR = VA – VB.
(b) Two bodies travelling in opposite direction ⇒ relative velocity, VR = VA + VB
→ Crossing of a river in shortest path :
(a) To cross the river in shortest path, it must be rowed with an angle, θ = sin-1(VWE/ VBW) perpendicular to flow of water.
(b) Velocity of boat with respect to earth,
VBE = \(\sqrt{\mathrm{V}_{\mathrm{BW}}^2-\mathrm{V}_{\mathrm{WE}}^2}\)
(c) Time taken to cross,
t = \(\frac{\text { width of river }(l)}{\text { velocity of boat w.r.t earth }}=\frac{l}{\mathrm{~V}_{\mathrm{BE}}}\)
→ Crossing the river in shortest time :
(a) Time taken to cross the river,
t = \(\frac{\text { width of river } l \text { l }}{\text { velocity of boat w.r.t water } \mathrm{V}_{\mathrm{BW}}}\)
(b) Resultant velocity of boat,
VR = \(\sqrt{\mathrm{V}_{\mathrm{BW}}^2+\mathrm{V}_{\mathrm{WE}}^2}\)
(c) Angle of resultant motion with
θ = tan-1(VWE/ VBE) down the stream
→ Dot product:
A̅ . B̅ = |A̅| . |B̅| cos θ
→ Let A̅ = x1 i̅ + y1 j̅ + z1 k̅ and
B̅ = x2 i̅ + y2 j̅ + z2 k̅ then
(a) A + B = (x1 + x2)i̅ + (y1 + y2)j̅ + (z1 + z2)k̅
(b) |A| = \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2+\mathrm{z}_1^2}\)
|B| = \(\sqrt{\mathbf{x}_2^2+\mathbf{y}_2^2+\mathbf{z}_2^2}\)
(c) A̅.B̅ = x1x2 + y1y2 + z1z2
→ In dot product i̅ i̅ = j̅ j̅ = k̅ k̅ = 1 i.e., dot product of heterogeneous vectors is unity.
→ In dot product i̅ . j̅ = j̅ . k̅ = k̅ . i̅ = 0
→ Projectiles thrown into the space with some angle ‘θ’ to the horizontal: Horizontal component (ux) = u cos θ. Which does not change.
→ Vertical component, Uy = u sin θ (This component changes with time)
→ Time of flight, (T) = \(\frac{2 u \sin \theta}{g}\)
H = \(\frac{u^2 \sin ^2 \theta}{2 g}\)
Range (R) = \(\frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{g}}\)
→ Velocity of projectile, v = \(\sqrt{v_x^2+v_y^2}\) where vx = ux = u cos θ and vy = u sin θ – gt
→ Angle of resultant velocity with horizontal, v.
α = tan-1\(\left[\frac{v_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}\right]\) where vy = u sin θ – gt and vx = u cos θ
→ In projectiles range R is same for complementary angles (θ and 90 – θ).
For θ = 45°, Range is maximum.
Rmax = \(\frac{\mathrm{u}^2}{\mathrm{~g}}\) corresponding to hmax = \(\frac{u^2}{4 \mathrm{~g}}\)
→ Relation between Rmax and hmax is
Rmax = 4hmax
→ For complimentary angles of projection,
h1 + h2 = \(\frac{u^2}{2 g}\);
Range, R = 4\(\sqrt{\mathrm{h}_1 \mathrm{~h}_2}\); Rmax = 2 (h1 + h2)
→ Horizontally projected projectiles: Time of flight, t = \(\sqrt{\frac{2 h}{g}}\)
→ Range, R = u × t = \(\sqrt{\frac{2 h}{g}}\)
→ Velocity of projectile after a time t is, v = \(\sqrt{v_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}\) where vx = ux = u and vy = gt
∴ v = \(\sqrt{\mathrm{u}^2+\mathrm{g}^2 \mathrm{t}^2}\)
→ Angle of resultant with x – axis,
α = tan-1\(\left[\frac{v_y}{v_x}\right]\) where vx = u and vy = gt
α = tan-1\(\left[\frac{\mathrm{gt}}{\mathrm{u}}\right]\)