# PC1431 Finals (1112) Hints

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### PC1431 – 11/12 Past Year Paper Hints

1. Refer to Tutorial 3 question 1

2. (a) Refer to Tutorial 2 question 3

(b) If T decreases, what happens to the motion in the vertical direction? Approach:

• When T decreases, the limiting static friction increases

• But as long as the applied force does not increase, the value of static friction itself will stay the same • Static friction is always equivalent to applied force as long as applied force is lower than limiting static

Therefore, nothing will happen.

3. Given:

m = 0.25kg r = 0.75m v = 25ms−1

(a) Question: Find M Approach:

• Centripetal force on m = gravitational force due to M Answer:

M = 21.2kg (b) Given: r0 = 0.65m

Question: Find change in M Approach:

• Conservation of angular momentum: Obtain v0: v0= 28.85ms−1 • Centripetal force on m = gravitational force due to M + ∆M Answer:

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4. Given:

(a) Question: Immediately after its release, α =? Approach:

• Rotational dynamics: τ = IPα

• IP can be found by using parallel axis theorem Answer: α = 2g 3R (b) Question: vcm=? Approach: • Conservation of energy

• Note: If you take the axis about point P , you need only consider rotational motion

mgR = 1 2IPω 2 • vcm= Rω Answer: v = r 4gR 3

V1= V0 p1= p0 p2= p0 8 (a) Question: V2=? Approach: • Adiabatic equation: P1V γ 1 = P2V γ 2 Answer: V2= 8 1 γV0

(b) Question: How much work is done? Approach:

• First law of thermodynamics: ∆U = ∆Q − ∆W • Adiabatic: ∆Q → 0

• Change in internal energy: ∆U = nCV∆T = −∆W • Finding ∆T

Adiabatic equations to express T2 in terms of T1: T1V1γ−1= T2V2γ−1 ∆T = (8γ1−1− 1) 1

nRP0V0

• Other equations that might be helpful in simplifying the expression: R = CP− CV, and CP CV = γ Answer: ∆W = 1 γ − 1p0V0 h 1 − 8(γ1−1) i

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6. Given:

m = 1kg TW = 0◦C TR= 100◦C

Given in the formula list: C = 4190 J kg−1 K−1 Question: ∆SW + ∆SR=?

Approach:

• Equation for entropy: ∆S =R dQ T • For heat reservoir:

Temperature is constant – integration not necessary: ∆S =∆Q T Find ∆Q: ∆Q = mC(TR− TW)

• For water:

Temperature is changing: Integration is necessary Express dQ in terms of dT : dQ = mCdT

Integrate with appropriate limits • Add up both entropy

∆S = +184 J K−1

7. Refer to question paper for details on question (a) Question: Find v at maximum compression

Approach:

• Inelastic collision: Conservation of momentum Answer:

v = m1v1+ m2v2 m1+ m2 (b) Question: Find maximum compression xmax

Approach:

• Conservation of energy1

• Substitute in the v obtained from part (a). Answer:

xmax= (v1− v2)

r m

1m2 k(m1+ m2)

(c) Question: Find velocity of each glider after m1 loses contact with the spring Approach:

• After m1 loses contact with spring, let: Velocity of m1 be vf1

Velocity of m2 be vf2

Therefore 2 unknowns → need 2 equations • Understand (and maybe explain) that:

– When m1 loses contact with the spring, this collision is elastic

– The spring gives back all its potential energy to the two gliders as kinetic energy

– Therefore the total kinetic energy of the two gliders before they hit the spring and after they separate from the spring are the same

– Total kinetic energy is conserved

1Inelastic collision merely states that kinetic energy is not conserved.

Some of the kinetic energy will be lost to the interactions between the two colliding masses

However, the energy given to the interaction is taken into account, we can use conservation of energy

In this collision, the two masses and the spring is an isolated system – therefore no energy is lost if we take into account potential energy of spring

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• Conservation of momentum • Conservation of kinetic energy2 Answer: v1f =  2m1 m1+ m2  v1+  m2− m1 m1+ m2  v2 v2f =  m1− m2 m1+ m2  v1+  2m2 m1+ m2  v2

(d) Question: Find the condition for exchanging velocities Approach:

• Just equate v1f = v2 and v2f = v1 and do the math

Condition: When m1= m2

8. Refer to question paper for details on the question (a) Question: If I = cM R2, find c

Approach:

• Add up moment of inertia of all the different parts Answer:

c = 5 12 (b) Wheel rolls without slipping down an incline of θ = 30◦

Question: a =?, f =? Approach:

• Rolling without slipping: static friction • Draw free body diagram3

• 2 unknown: 2 equations Linear equation of motion

M g sin θ − fs= M a Rotational equation of motion

fsR = Iα • Rolling without slipping: a = Rα

• Solve for fs and a Answer:

a = 3.46m s−1 fs= 34.62N

(c) Rope is pulled to the right with T = 150N. It rolls without slipping Question: a =?

Approach:

2Or approach speed = separation speed

3If you’re not sure of the direction of static friction, you can assume a direction – if your direction is wrong, your answer will be negative. In this question however, it should be pretty obvious that it points up the slope

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• Similar to the above

• Still 2 unknown (fsand a): 2 equations Linear equation of motion

T − fs= M a Rotation equation of motion

fsR − T (0.2R) = Iα • Rolling without slipping: a = Rα

a = 3.53m s−1 (d) Given:

Question: Will the wheel move along the plane? Approach:

• Problem I: Should we consider fsor fk?

As our job is to determine if the wheel will move, we will be looking at fs. We only look at fk if the wheel is already moving

• Problem II: Direction4of f s

As before, can be assumed – if you assumed the wrong direction, you’d get a negative sign • Problem III: 3 unknowns: T , fs and a

T and fschange according to the situation

fswill only be equal to the limiting static friction if the wheel moves • Therefore, we shall let a = 0, and check what fsis.

– The fs calculated here will the friction necessary to keep the wheel from moving – If fs< µsN (limiting static friction), we know that it will not move

– If the needed fs> µsN , then it will move • When a = 0, 2 unknowns (T , fs): 2 equations

Linear equation of motion

M g sin θ − T + fs= 0 Rotational equation of motion

T (0.2R) − fsR = 0

4We chose this direction because only T and f

sprovide torque. Therefore, if we assume a = 0, as explained in the point below, fsmust oppose T

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• Solve to find fs

You will find fs< µsN Answer:

No it will not move down because calculated static friction is lower than the limiting static friction.

9. Given: Heat engine cycle

(a) Question: W in one cycle Answer:

W = (p2− p1)(V2− V1) (b) Question: QH flowing in for one cycle

Approach:

• Know that heat enters the gas at a → b and b → c (Alternatively, know that QH is positive Q) • Q = nCV∆T for isochoric and Q = nCP∆T for isobaric

• Use ideal gas law to express n and T in terms of P , V and R Answer: QH= CV R (p2− p1)V1+ CP R (V2− V1)p2 (c) Question: QC flowing out for one cycle

Approach: Same as above Answer: |QC| = CP R (V2− V1)p1+ CV R (p2− p1)V2 (d) Question: QH+ QC− W in one cycle

0 as QH+ QC− W = ∆U , and this is a cyclic process, therefore ∆U = 0 (e) Given: p2= αp1 and V2= αV1, where α > 1

Question e =? Approach:

• For engine, equation for e = W QH

= 1 −|QC| QH

• Use equations and values obtained/given above to determine final answer • Know: CP = CV + R

e = (α + 1)(γ − 1) 1 + γα (f) Given: Cycle is reversed and made into a refrigerator

Question: Find the coefficient of performance K Approach:

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• Similar to the above • Know: K = QC W Answer: K = γ + α (γ − 1)(α − 1)

References

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