Slide 1 : PROF. ARDAMAN SIDHU
HKSIS
PRESENTS
LECTURES IN PHYSICS
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How to solvePHYSICS NUMERICALS : How to solvePHYSICS NUMERICALS QUICK AND EASY SOLUTIONS
Slide 3 : The technique Prof. Ardaman Sidhu
For all your physics problem
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Slide 4 : The step–1 It is the visual simplification of the problem. SKETCHING This helps a student in, in-depth understanding of the problem. Note
60% of the problem gets solved, during the sketching of the problem. Read the whole problem and make sure you understand it.
Then read it again and then sketch Draw a diagram and choose coordinate axes.
Slide 5 : DIVIDE & RULE It is a well known fact, that ‘divide & rule’ though bitter, helps in better and successful ruling. Similarly, a student should be aware of the principle of ‘how to divide & rule’ the problem. During the application of this step, we
split the problem into segments or cases. The division shall happen in such a way, that one case contains given data and other/s contain/s the required terms. The step–2
Slide 6 : (a) Direct data: DATA NOTING The step–3 This step is also called ‘Heart of the problem’ . Let us see, how a data has to be noted systematically. Noting of direct data includes, the terms given in the problem. During data noting, you shall ensure that data in both the cases is synchronized. This method, balances and sets the clarity in data. (b) Indirect data: This data can be noted from the conditions derived from the problem or known conditions, like constants.
Slide 7 : Note: During data noting ensure…… (i) Required term is identified by using a question mark (?) (ii) New terms selected, shall minimize the number of variables. (iii) The answer is always expressed in terms of the given data or known constants.
Slide 8 : The step–4 FORMULA SELECTION Formula selection can be considered as the ‘Problem solving Strategic plan’. Select an appropriate formula
That helps us reach the final goal – the answer. Formula can be selected, by checking the terms present in the data noted. Ensure…..
Formula contains the known terms and required term. What physics applies here?
Plan an approach to a solution.
Slide 9 : The step–5 GENERATION OF EQUATIONS How to generate the equations ? If it is a simple case
Then generally we have to substitute terms in the formula. If more than one case ….
Apply the formula and generate the equations. Rule for substitution
Make sure that all the terms are present in a same system of unit.
Slide 10 : The step–6 ANALYSIS OF EQUATIONS Observe (or) make note of the number of terms present in the equations. Let us see, how to analyze the equations.. 1) Identify the concerned terms: 2) Unconcerned terms: These are the terms that are given and are needed. These are the terms which are assumed or introduced. The ‘indirect data’ contains the unconcerned terms. The ‘direct data’ contains the concerned terms.
Slide 11 : The step–7 ELIMINATIONS OF UNCONCERNED TERMS We know,
That the answer should be expressed only in terms of given data. Hence…
There is a need to eliminate the unconcerned terms. Can be removed by various mathematical manipulations. The common method to eliminating the unconcerned terms
Divide one equation with the other.
CROSS CHECK : CROSS CHECK Calculate the solution
Round it to the appropriate number of significant figures.
Look at the result – is it reasonable? Does it agree with a rough estimate?
Check the units again.
Slide 13 : Step – 7:Elimination of unconcerned terms Step – 6: Analysis of equations Step – 5 : Generation of equations Step – 4: Selection of formula Step – 3: Data noting Problem Solving technique – A Summary Step – 2 : Divide and Rule Step – I: Sketching NO EXPECTATIONS BUT ACTION:
Do not expect any idea (or) plan. If you get an idea or plan, it is a Bonus.
When nothing is expected in life…
Then every thing we get is a bonus and gift.
Slide 14 : A body moving with a uniform acceleration has a velocity ‘U’ at point A and velocity ‘V’ at point C. Find the velocity at B, the midpoint of AC. (i) Sketching U V v' v' =? Case –1(AC) Case –2 (AB) u1 v1 a1 u2 = v2 = a2 = = U = a U v'=? a Sol. Let a and v' be the uniform acceleration and velocity at B, Let us, use the following scientific steps of problem solving, to solve the above problem. (ii) Divide and Rule (iii)
Data noting (b) Indirect data (a) direct data s1 t1 s2 = t2= = 2x x t' Now, we can introduce two terms either ‘s’ or ‘t’ so .. = t
Slide 15 : v2 – u2 = 2as (iv) Selection of the formula: (v) Generation of equations: v2 – u2 = 2 ? a ? 2x = 4ax …(1) v'2 -u2 = 2 ? a ? x = 2ax …(2) 1. A body moving with a uniform acceleration has a velocity ‘U’ at point A and velocity ‘V’ at point C. Find the velocity at B, the midpoint of AC.
Slide 16 : (vi) Analysis of equations: The two equations connect 5 terms v, u,a,x and v’ out of which ___,____ and v’
____ are concerned and _____ and ___ are unconcerned. (vii) Elimination of unconcerned Eliminate the unconcerned terms by dividing equation (2) with (1) v u 2
Slide 17 : tAC = Case (i) Case (ii) AB AC t1 t2 = a1 a2 = u1 u2 = S = ut + ½ at2 2) A freely falling body took t second to come down from top of a tower. Time taken to cover half the height of the tower is _________________ tAB = = t t’ = ? = g g = 0 0 = h = h/2 t t’ = ? h h/2 Let’s solve the next problem… S1 S2 Applying for both the cases we get, Formula: Sol:
Slide 18 : C
• B
• S1=K m v = 0 BC =? CASE I CASE II u/2 u AB BC S1 = K S2 = BC = K’=? u u1 = u2 = u/2 = u/2 v1 v2 = 0 = – a a1 a2 = – a v2 – u 2 = 2aS (u/2)2 – u2 = 2? – a ? K 2aK –––––––––– (1) 02 – (u/2)2 = 2 ? – a ? K’ 2 a K’ ––––––– (2) 3) A body is moving such that it covers a distance of K m during which its velocity gets halved. Then, what is the further distance it travels before coming to rest? S2=? Let’s solve the next problem… Applying for both the cases we get, Formula: Sol: A
Slide 19 : 4) A body falling from rest has acquired a velocity v after it falls through a distance h. The distance it has to fall down further for its velocity to double is _________ times h. u=0 h 2v v CASE I CASE II = h = 0 = v = g S2 = u2 = a2 = v2 – 02 = 2gh ? (2v)2 – v2 = 2gh’ ? v2 = 2gh 3v2 = 2gh’ Dividing eq. 2 with 1 we get, S1 u1 a1 BC =h’=? v = 2v g BC =h’=__h ? Let’s solve the next problem… Applying for both the cases we get, Formula: v1 v2 (AB) (BC) Sol:
Slide 20 : C
• u 2 u 4 u Case (i) Case (ii) AB AC = tAB = t t2 = = 2u v2 = = u u2 = = a a2 = We know that 2u = u + at ? u = at ––––(1) 4u = u + at’ ? 3u = at’ –––(2) 5) A body moving with uniform acceleration has its velocity doubled in t second. The velocity quadruples in ___________second. v = u + at Dividing equation (2) with (1) we get, t1 v1 u1 a1 tAC =t’ = ? 4u u a A
• B
• tAB = t tAC=? Let’s solve the next problem… Applying for both the cases we get, Formula: Sol:
Slide 21 : A B C x BC = x =? ? u2=75m/s Case (i) Case (ii) AC? BC? = 300 – x S2 = = 0 u2 = = g a2 = = v v2 = = t t2 = S1 u1 a1 v1 t1 x 75 m/s – g t 6) From a tower of height 300m, a body is allowed to fall freely and simultaneously, a body from the bottom of the tower is projected up with a velocity 75 m/sec. What is the height from the ground level, where the two bodies meet? ( Take g = 9.8m/s2) ? u1 = 0 Let they meet at a distance x from the ground level v1 300 – x = x = Adding (1) and (2) we get, Substitute t = 4 in eq(2) x = (75?4) – 1/2 ? (9.8) ? 42 = 221.6m Let’s solve the next problem… Applying for both the cases we get, Formula: Sol: 300 – x
Slide 22 : 7) A police man sitting on his motor cycle observes a thief moving uniformly in a car at 30 m/s and starts from rest simultaneously chasing him, with an acceleration of 3m/s2. When and where he is going to catch the thief? u=30m/sec a=3 m/sec2 AB = PQ = ? = 30m/s = t = ? u2 = = 0 a2 t2 = Formula: Police u1 t1 a1 S1 = x =? 0 = 3m/s2 t = ? S = ut + 1/2 a t2 By equating (1) and (2) we get t = ? S2= x =? Applying for both the cases we get, Sol:
Slide 23 : 8) Two cars are travelling towards a junction along two different intersecting roads. One of them is 200m away and is moving uniformly at 10m/s, the other is located 100m away starts from rest and accelerates uniformly at 2 m/s2 towards the junction. The interval between the moments when the cars pass the intersection is ______ u = 10m/s S1 = 200m a= 2m/s2 CAR–1 CAR–2 S1 u1 a1 = 200m = 10m/s = 0 CA BA S2 = u2 = a2 = 100m 0 2m/s2 t1 = t t 2 = t’ Formula: Apply for both cases we get, S2 = 100m u= 0 J A C B Let’s solve the next problem… Sol:
Slide 24 : h v/3 v/2 9) A stone projected so as to reach a height h passes P and Q with velocities v/2 and v/3. The distance between the points is _______ h , where v is the initial velocity with which the body is thrown . CASE–1 CASE–2 v u 2 v1 v2 h S2 = X =? a1 = a2 Applying for both the cases we get, u = v S1 = u 1 = = v/2 = 0 = v/3 –g = – g Let’s solve the next problem… Sol: (AB) (PQ) Formula: (2) (1) X=?
Slide 25 : 10) Find the displacement by a body which covers an arc of a circle which subtends an angle of 120° at the centre of the circle if radius is r. =AB =? Sin 600= Now, AB = 2BC ? AB = 2? rsin 60 Displacement AB = AC + BC = 2AC or 2BC How to get BC = ? In ? COB Sol:
WONDERFUL WORLD OF PHYSICS L2/AS 3rd,June,Wed : WONDERFUL WORLD OF PHYSICS L2/AS 3rd,June,Wed
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Email ………………. hksidhuinstitute@gmail.com