PC1431 Finals (1112) Hints

August 28, 2017 | Author: Leonard | Category: Collision, Rotation Around A Fixed Axis, Mechanics, Chemistry, Classical Mechanics
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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 friction Answer: Therefore, nothing will happen.

3. Given:

m = 0.25kg

r = 0.75m

(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 v 0 : v 0 = 28.85ms−1 • Centripetal force on m = gravitational force due to M + ∆M Answer: ∆M = 11.4kg


v = 25ms−1

4. Given:

(a) Question: Immediately after its release, α =? Approach: • Rotational dynamics: τ = IP α • 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 IP ω 2 2

• vcm = Rω Answer: r v=

4gR 3

5. Given: Adiabatic expansion V1 = V0

p1 = p0

p2 =

p0 8

(a) Question: V2 =? Approach: • Adiabatic equation: P1 V1γ = P2 V2γ Answer: 1

V2 = 8 γ V0 (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 : T1 V1γ−1 = T2 V2γ−1 1

∆T = (8 γ −1 − 1)

1 P0 V0 nR

• Other equations that might be helpful in simplifying the expression: R = CP − CV , and Answer: ∆W =

h i 1 1 p0 V0 1 − 8( γ −1) γ−1 2

CP =γ CV

6. Given: TW = 0◦ C

m = 1kg

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: ∆Q Temperature is constant – integration not necessary: ∆S = 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 Answer: ∆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=

m1 v1 + m2 v2 m1 + m2

(b) Question: Find maximum compression xmax Approach: • Conservation of energy1 • Substitute in the v obtained from part (a). Answer: xmax = (v1 − v2 )


m1 m2 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 1 Inelastic 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


• Conservation of momentum • Conservation of kinetic energy2 Answer:  v1f =

2m1 m1 + m2

 v1 +

m2 − m1 m1 + m2


v2 f =

m1 − m2 m1 + m2

 v1 +

2m2 m1 + m2


(d) Question: Find the condition for exchanging velocities Approach: • Just equate v1f = v2 and v2f = v1 and do the math Answer: 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 fs R = 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: 2 Or

approach speed = separation speed 3 If 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


• Similar to the above

• Still 2 unknown (fs and a): 2 equations Linear equation of motion T − fs = M a Rotation equation of motion fs R − T (0.2R) = Iα • Rolling without slipping: a = Rα • Solve for a Answer: a = 3.53m s−1 (d) Given:

Question: Will the wheel move along the plane? Approach: • Problem I: Should we consider fs or 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: Direction4 of fs 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 fs change according to the situation fs will only be equal to the limiting static friction if the wheel moves • Therefore, we shall let a = 0, and check what fs is. – The fs calculated here will the friction necessary to keep the wheel from moving – If fs < µs N (limiting static friction), we know that it will not move – If the needed fs > µs N , 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) − fs R = 0 4 We chose this direction because only T and f provide torque. Therefore, if we assume a = 0, as explained in the point below, f must s s oppose T


• Solve to find fs You will find fs < µs N 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 =

CP CV (p2 − p1 )V1 + (V2 − V1 )p2 R R

(c) Question: QC flowing out for one cycle Approach: Same as above Answer: |QC | =

CP CV (V2 − V1 )p1 + (p2 − p1 )V2 R R

(d) Question: QH + QC − W in one cycle Answer: 0 as QH + QC − W = ∆U , and this is a cyclic process, therefore ∆U = 0 (e) Given: p2 = αp1 and V2 = αV1 , Question e =? Approach: • For engine, equation for e =

where α > 1

W |QC | =1− QH QH

• Use equations and values obtained/given above to determine final answer • Know: CP = CV + R Answer: 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: 6

• Similar to the above QC • Know: K = W Answer: K=


γ+α (γ − 1)(α − 1)

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