Points:
- Amount,length,time,luminous intensity, mass, temperature, current are the fundamental physical quantities.
- Same units in every term on either side of =
- When round determining digit is 5 and the truncated digits are 0 then the number is rounded up if the number to remain is odd and truncated if the number to remain is even.
- For ± least decimal places, for */ least significant figure
- If something else pushes or pulls an object, its velocity will change
Trivia:
- Df(x) = f'(x)
- You can move vectors
- Motion of object with respect to another object is relative motion
- When relative velocity is towards the reference point it is called velocity of approach
- Forces are either contact or non-contact
Non-formulae:
- a = am ± Δa {Δa = Σ|am-ai|}
- nVSD = (n-1)MSD
- LC = VC = 1MSD-1VSD = (1/n)MSD
- LC = pitch/number of divisions
- |A+B| = sqrt(|A|2 + |B|2 + 2|A||B|cosθ)
- Angle of A+B with A = tan-1(Bsinθ/(A + Bcosθ))
- A•B = |A||B|cosθ = AxBx + AyBy + AzBz
- AxB = det(i j k , Ax Ay Az, Bx By Bz) = |A||B|sinθ * direction
- v = u + at
- s = ut + at2/2
- v2- u2 = 2as
- sn = u + a(2n-1)/2
- H = (usinθ)2/2g = uy2/2g
- R = u2sin2θ/g = 4Hcotθ
- R (for horizontal projectile) = uxuy/g = uxsqrt(H/2g)
- Minimum speed to hit (x,y) = umin = sqrt(g(y+sqrt(x2 + y2)))
- Minimum angle = tan-1[umin2/gx]
Conditional:
Elastic collision with wall
If a wall is in the path of a projectile at a distance x from the starting point and the collision is elastic then:
- tf and H remain the same , distance covered after the collision will be in opposite x direction and same y direction as distance covered before collision
- If x ≥ R/2 landing spot is at distance 2x-R from the launch point in the same x direction as initial velocity
- If x ≤ R/2 landing spot is at distance R-2x from the launch point in the opposite x direction to initial velocity
Projectile on inclined plane
For up the plane:
- The angle of inclination is α
- The acute angle of projectile be θ
- tf = 2usinθ/gcosα
- R = u2[sin(2θ+α)-sinα]/gcos2α
- R is max when θ = (π/4) - (α/2) and, Rmax = u2/g(1+sinα)
- Max distance from inclined plane = u2sin2(θ)/2gcosα
For down the plane replace α with -α to get the corrected expressions
Man swimming in river
- Vman = Vman|river + Vriver {v = |Vman|river|}
- Time taken to cross river, t = d/vsinθ , θ = angle between Vman|river and bank of the river
- Drift = (vcosθ + u)*t (u=|Vriver|)
- For drift to be zero , cosθ = -u/v , t = d/sqrt(v2 - u2)
- If v<u minimum non-zero drift is when cosθ = -v/u
- For swimming in a specific direction,
Angle of Vman|river with direction of swim be α
vsinα = usinθ , θ is angle of direction of swim with river bank
Minimum distance between 2 moving bodies:
Initial distance = L
Angle made between Vrel and initial displacement = θ
Angles made by the moving bodies be α & β and velocities be Va , Vb
- Minimum distance = Lsinθ
- |Vrel| = sqrt(|Va|^2 + |Vb|^2 +2|Va||Vb|cos(α + β))
- Time to reach minimum separation is Lcosθ/|Vrel|
For collision:
- |Va|sinα = |Vb|sinβ
- Time taken to collide = L/(|Va|α + |Vb|cosβ)
Formulae:
- Zero error = -zero correction
- Length = reading + zero correction
- (1+x)n = 1 + nx + n(n-1)x/2! + n(n-1)(n-2)x/3! + ... , |x|<1
- a/sin(A) = b/sin(B) = c/sin(C)
- A = |A|*dir(A)
- |A| = sqrt(Ax2 + Ay2 + Az2)
- DA|B = DA - DB
- tf = 2(usinθ)/g
- td = sqrt(H/2g)
- For man in rain to remain dry, he should hold his umbrella at an angle of θ = tan^-1(Vm/Vr)
- a = F/m
Tips:
- Use dot product to find angle between vectors
- Verify if graph given is trajectory or distance-time graph
- For projectile at a height, find tf using s = ut + at2/2
- For finding shortest distance, consider relative motion
- For collisions, the trajectories should intersect and at the point of intersection the times must be the same
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