Past question on Wein's Law and Stefan's Law

 

Mark scheme for questions on circular motion

 

1.       (i)      (v = 2πr/t) t = 2π60/0.26 = 1450 s

Correct answer is 1449.96 hence allow 1.4 × 103 
Do not allow a bare 1.5 × 103

B1

  (ii)   correct substitution into F = mv2/r: eg F = (9.7 × 103 × 0.262)/60

C1

          F = 10.9 N

Allow 11 N

A1

[3]

 

  2.     (i)      THREE correct arrows at A, B and C all pointing towards
the centre (judged by eye)

Ignore starting point of arrow

B1

  (ii)   1.      Greatest reaction force is at C

This is a mandatory M mark. The second mark cannot be gained unless this is scored.

M1

         because it supports weight of sock AND provides the required
upward resultant (centripetal) force (WTTE)

Any indication that candidates think that the centripetal force is a third force loses this second and possibly the next mark. They must make correct reference to the resultant force that provides the required centripetal force/acceleration.

A1

2.      Least at A because sock’s weight provides part of the required
downward resultant (centripetal) force (WTTE)

Allow answers using the equation F = mv2/r
such as Nc – mg (at C) = centripetal force OR mv2/r
OR mg +NA (at A) = centripetal force OR mv2/r

B1

[4]

 

  3.     (i)      At top of loop, the centripetal force = mv2 / r (1)
= mg (1)
Thus speed at top v = √gr
= √9.81 × 9.17 / 2 (1)
= 6.7 m s–1                                                                                                         3

          or use an energy argument KE on entry = PE gained + KE at top

          (1 mark for idea, 1 mark for correct substitution in appropriate formula
and 1 mark for correct calculation of 6.7 m s–1)

          (NOTE The unexplained use of v2 = u2 + 2gs can only score a
maximum of 1 mark)

(ii)     Kinetic Energy at top = ½ × 86 × 6.72 (1)
= 1935 J (1)
Potential Energy at top = 86 × 9.81 × 9.17 (1)
= 7740 J (1)                                                                                                       4

  (iii)  Kinetic Energy on entry = ½ × 86 × 152
= 9675 J (1)
Sum of energies at top = 1935 + 7740
= 9675 J .......... QED (1)                                                                                    2

  (iv)  Any reference to loss of contact / centripetal force or wtte (1)
Comment on the consequences of taking off vertically or wtte (1)                         2

          Enacting the suggestion could result in disaster
At the point A in the loop, the velocity vector is purely vertical
Therefore there is no horizontal component of velocity
So no matter how fast the cyclist is travelling he will only be
projected vertically
And come (crashing?) down on the same point where he left off
The best that could happen ( with some skill ) is to return back along same path

[11]

 

  4.     (a)     B = F/Il with symbols explained or appropriate statement in words; (1)
explicit reference to I and B at right angles/define from F = BQv etc (1)                2

  (b)   (i)      arrow towards centre of circle                                                                    1

(ii)     field out of paper; Fleming’s L.H. rule/moving protons act as
conventional current                                                                                 2

  (iii)  F = Bev allow BQv                                                                                   1

  (iv)  F = mv2/r; Bev = mv2/r; (2)
B = mv/er = 1.67 × 10–27 × 1.5 × 107/(1.6 × 10–19 × 60); = 0.0026; T (3)      5

allow Wb m–2

  (v)   the field must be doubled; (1)
B ∞ v (as m, e and r are fixed)/an increased force is required
to maintain the same radius (1)                                                                  2

[13]

 

  5.     (a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                              2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                  1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                   2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown                  2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

         Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                                  5

  (b)   (i)      k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)                                                                                 2

  (ii)   f = (1 /2π (k/m)0.5 (0)
= (1 /2π (2.0 × 106 / 9.5 × 105)0.5 (1)
= 0.23 Hz (1)                                                                                            2

  (iii)  If wind energy causes this frequency in the structure, the
amplitude increases / resonance occurs / or explanation of
resonance / ref. to natural frequency (1)
e.g. damping is necessary / mass change to shift resonant
frequency / change spring constant (1)                                                       2

[18]

 

  6.     the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                             3

MAXIMUM 3

[3]

 

Questions on Speed, Velocity and Acceleration

 

3.5 Newton’s laws of motion MS

 Q1              B

Q2              B

Q3              D

Q4              B

Q5              D

Q6              D

M7.          (a)     kinetic energy is not conserved (1)

(or velocity of approach equals velocity of separation)

1

(b)     (i)      (use of p = mv gives) p = 4.5 × 10^–2 × 60 (1)

 = 2.7kg m s^–1 (1)

8.       (i)      Force is proportional to the rate of change of momentum
(QWC This mark can only be scored if momentum is spelled correctly)

Allow “equal” instead of proportional, allow “change in momentum over time” (WTTE)
Do not allow F = ma or in words

B1

  (ii)   When one body exerts a force on another the other body exerts an equal
(in magnitude) and opposite (in direction) force on the first body (WTTE)

Must refer to two bodies.
Do not allow a bare “Action and reaction are equal and opposite”.

B1

[2]

  9.     kinetic energy is the energy a body possesses by virtue of its speed (1)
s an energy it is a measure of force × distance (1)
the rate of change of momentum defines force (1)
momentum is therefore a measure of force × time (1)                                                 4

          Other possible answers will score a maximum of 3 unless the
force × distance relationship is given for kinetic energy and the
force × time relationship is given for momentum

          momentum is always conserved in a collision (in the absence of
external forces) (1)
but kinetic energy may be lost – with qualification of what happens (1)
kinetic energy is proportional to v2 but momentum is proportional to v (1)
kinetic energy is a scalar; momentum is a vector (1)

[4]

 

  10.   (a)     (i)      Mass × velocity/mv with symbols defined                                                  1

(ii)     0 = mAvA ± mBvB or mAvA = mBvB (1)
vA/vB = ± mB/mA (1)                                                                                    2
max 1 mark for final expression without line 1

  (b)   (i)      vA = (10/5 =) 2.0 (m s–1) and vB = (10/10 =) 1.0 (m s–1)                            1

(ii)     t1 = 3.0/2.0 = 1.5 (s) ecf b(i)                                                                       1

(iii)     x = 2.1 – 1.0 × 1.5 = 0.6 (m)                                                                      1

(iv)    v = vB + (5/50)vA = 1.0 + 0.2 (= 1.2 m s–1)                                                 1

(v)     t2 = t1 + 0.6/1.2 = 2.0 (s)                                                                             1

(vi)    At collision the container (and fragments) stop (1)
By conservation of momentum, total momentum is still zero/AW (1)        2

(vii)   straight lines from (0, 0) to (1.5, 0); (1.5, 0) to (2.0, 0.1);
(x, 0.1) for all x > 2                                                                                     3

[13]

Attenuation of X Rays

 

Question on Big Bang

 

1st part only

Elastic and Inelastic Collisions

 

Hubble Question - balloon