4.6 Mechanics and Materials - Momentum 2 – Mark schemes

Q1. (a) Attempt to determine area under graph or statement that area under needed or 0.5

× 15 × 10−3 x 66 ✔

0.495 (N s)✔ condone power of 10 error

2

(b) Momentum before = 0.045 × 7.1 = 0.320 (N s) down✔

Momentum after = 0.045 × 3.9 = 0.175 (N s) upwards✔

Change = 0.495 (N s) ✔ 3

(c) Initial KE on impact = 0.5 × 0.045 × 7.12 = 1.13 (J) or Ke after impact = 0.342 (J)✔

Fractional change ke after / ke before = 0.30✔

Use of their fractional change cubed✔

Percentage change after 3 bounces = 0.33 × 100 (%) = 2.7%✔ 4

(d) As ball falls momentum of ball toward the Earth (always) = momentum of Earth toward the ball✔

On impact the momentum of both ball and Earth become zero✔

After impact momentum of ball away from Earth = momentum of Earth in opposite direction✔

3 [12]

Q2. (a) Total mass of spacecraft = 3050 kg

Change in PE =

1.9 × 1011(J)

2 sf condone errors in powers of 10 and incorrect mass for payload Allow if some sensible working

4

(b) Chemical combustion of propellant / fuel or gases produced at high pressure

Gas is expelled / expands through nozzle

Change in momentum of gases escaping

equal and opposite change in momentum of the spacecraft

Thrust = rate of change of change in momentum Max 3 N3 in terms of forces worth 1

3

(c) 0.031(4) (m s-2) 1

(d) Use of rocket equation

v = 1200 ln 996 (m s–1)

Condone 1000 (m s–1) 3

(e) (i) Use of correct mass 108 kg

0.0198 N Allow incorrect powers of 10 and mass

3

(ii) Use of v =

Correct substitution v =

0.86 (m s-1) Recognisable mass – condone incorrect power of 10

3

(iii) Impulse = 25 N × 4.8 = 120 N s

(120 = 108 v so) Velocity = 1.1 m s-1

Clear conclusion

ie explanation/comparison of calculated velocity with escape velocity from (e)(ii)

May use F = ma approach 3

[20]

Q3. (a) use of E = mgh (1)

2680 J (1) 2

(b) use of v2 = 2 as (1)

9.1 m s-1 (1)

2

(c) increases the time taken for the athlete to come to rest/reduced deceleration

force = mass × acceleration/mass × change in velocity/time (1)

or momentum argument

or energy argument (1) 2

[6]

Q4. (a) (i) (change in momentum of A) = – (1) 25 × 103 (1)

kg m s–1 (or N s) (1)

(ii) (change in momentum of B) = 25 × 103 kg m s–1 (1) 4

(b)

initial vel/m s–1 final vel/m s–1 initial k.e./J final k.e./J

truck A 2.5 1.25 62500 15600

truck B 0.67 1.5 6730 33750

(1) (1) (1) (1) 4

(c) not elastic (1) because kinetic energy not conserved (1) kinetic energy is greater before the collision (or less after) (1) [or justified by correct calculation]

3 [11]

Q5. (a) (i) 1.5 × 104/1.46 × 104 (14550) (kg m s–1)

B1 1

(ii) correct substitution into t = (mv – mu)/F or (t =) 14550 ÷ 6.1 × 103 seen (condone power of 10 error) [ecf from part ai]

C1

t = 2.39 s [ecf from part ai]

C1

correct substitution into s = (u + v)t/2 or (s =) 15 × 2.39 ÷ 2 seen (condone incorrect value for a calculated t in substitution)

C1

s = 17.9 to 18 m [ecf from part ai]

A1 4

(b) (braking distance) increases/‘longer’ to stop

M0

greater mass

A1

more momentum/more time with rate of change of momentum equation

A1

same velocity change over longer time means greater distance /appropriate equation of motion with s as subject and longer time

A1

when using s = vt must identify v as average

or (braking distance) increases/‘longer’ to stop

M0

greater mass

A1

more ke (to convert)/more work to be done

A1

more work done by (same) force means greater distance/ appropriate equation with s as subject

A1

or (braking distance) increases/‘longer’ to stop

M0

greater mass

A1

smaller acceleration

A1

smaller acceleration for (same) change in velocity means greater distance/appropriate equation of motion with s as subject

A1

3 [8]

Q6. (a) (i) clear working:

attempts to evaluate area under line (12 – 13 squares × 0.1 N s) or ½ × 106(105) × 24 × 10–3 seen

B1

1.260 – 1.272 (N s) 1.2 – 1.3 for square counting

B1

(ii) area under graph/impulse = Δmv

B1

22.3 (22.8) (m s–1)

B1 4

(b) use of Pythagoras or tan–1 (6.1/22.3)

B1

or or tan–1 (6.1/20)

22.8 – 23.8(m s–1) 20.9 or 21(m s–1)

B1

14.9 – 15.5 (°) (15 – 16 (°)) 17/18(°)

B1 3

[7]

Q7. (a) force = rate of change of momentum (1)

1

(b) (i) area under graph represents impulse or change in momentum (1) 1

(ii) suitable method to estimate area under graph (1)(1)

[eg counting squares: 20 to 23 squares (1) each of area 25 × 10–3 × 20 = 0.5 (N s) (1) or approximate triangle etc (1) ½ × 250 × 10–3 × 90 (1)]

gives impulse = 11 ± 1 (1)

N s (or kg m s–1) (1) 4

(iii) use of impulse = Δ(mv) (1)

Δp = mv – (–mu) = m(v + u) or 11 = 0.42 (v + 10) (1)

giving 0.42 v = 6.8 and v = 16 (m s–1) (impulse = 12 gives 19 m s–1) (1)

answer to 2 sf only (1) 4

(c) final speed would be lower (1)

any two of the following points (1)(1)

• initial momentum would be greater [or greater u must be reversed]

• change in momentum [or velocity] is the same [or larger F acts for shorter t]

• initial and final momenta are (usually) in opposite directions

• initial and final momenta may be in same direction if initial speed is sufficiently high

[alternatively]

final speed = – initial speed (1)

gives final speed v = (26 ± 3) – initial speed u (1)

consequence is

• v is in opposite direction to u when u < 26

• v is in same direction as u when u > 26

• v is zero (ball stationary) when u = 26

any one of these bullet points (1) 3

[13]

Q8. (a) The candidate’s writing should be legible and the spelling,

punctuation and grammar should be sufficiently accurate for the meaning to be clear.

The candidate’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and

style of writing is appropriate to answer the question.

The candidate states that momentum is conserved, supported by reasoning to explain why the conditions required for momentum conservation are satisfied in this case.

The candidate also gives a statement that total energy is conserved, giving detailed consideration of the energy conversions which take place, described in the correct sequence, when there is an explosion on a body that is already moving.

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate.

The candidate states that momentum is conserved, but the reasoning is much more limited.

and/or

There is a statement that (total) energy is conserved, with basic understanding that some energy is released by the explosion.

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.

The candidate indicates that either momentum or energy is conserved, or that both are conserved. There are very limited attempts to explain either of them.

The explanation expected in a competent answer should include a coherent selection of the following points concerning the physical principles involved and their consequences in this case.

Momentum

• momentum is conserved because there are no external forces acting on the overall system (probe plus capsule) – or because it’s free space

• they are moving in free space and are therefore so far from large masses that gravitational forces are negligible

• during the explosion, there are equal and opposite forces acting between the probe and the capsule

• these are internal forces that act within the overall system

• because momentum has to be conserved, and it is a vector, the capsule must move along the original line of movement after the explosion

Energy

• total energy is always conserved in any physical process because energy can be neither created nor destroyed

• however, energy may be converted from one form to another

• the probe is already moving and has kinetic energy

• in the explosion, some chemical energy is converted into kinetic energy (and some energy is lost in heating the surroundings)

• the system of probe and capsule has more kinetic energy than the probe had originally, because some kinetic energy is released by the explosion

max 6

(b) (i) conservation of momentum gives (500 × 160) = 150 v + (350 × 240) (1) from which v = (−)26(.7) (m s−1) (1)

direction: opposite horizontal direction to larger fragment [or to the left, or backwards] (1)

3

(ii) initial Ek = ½ × 500 × 1602 (1) (= 6.40 × 106 J)

final Ek = (½ × 350 × 2402) + (½ × 150 × 26.72) (1) (= 1.01 × 107 J)

energy released by explosion = final Ek − initial Ek (1)

= 3.7 × 106 (J) (1) 4

[13]

Q9. (a) force is equal to (or proportional to) rate of change of momentum

[or impulse = force × time = change of momentum]

[Answer should not be in symbols unless all the symbols are explained] 1

(b) (i) use of mgΔh = ½ mv2 gives v = = ( = 5.60 m s–1) 1

(ii) momentum per second (= 0.30 × 5.60) = 1.68 (Ns) 1

(iii) mass of sand falling in 10s = (0.30 × 10) (= 3.00 kg)

force due to arriving sand = momentum arriving per second = 1.68(N)

equivalent mass reading =

(= 0.17 kg)

so balance reading is 3.00 + 0.65 + 0.17 ( = 3.82 kg ) 3

(c) horizontal lines up to 5 s and beyond 25 s

line of constant positive gradient between 5 s and 25 s

(near) vertical steps up at 5 s and down at 25 s 3

[9]

Q10. (a) 32.5 × 0.156 or 65 × 0.156: any mass × velocity

C1

10.07/10.1/10 ignore sf

A1

kg ms–1 (accept Ns)

B1 3

(b) their a/3.80 (× 10–3) ignore power of 10 error

C1

2670 (N) ecf

A1 2

[5]

Q11. (a) smooth curve with a maximum value shown

B1 condone non-zero at start and finish

gradient fairly constant or slight increase for half time

B1

falls gradually on second half of swing

B1 oscillations score zero

2 max

(b) impulse is product of force and time

B1 clear reference to impulse

prolonging the time (of contact) increases momentum / velocity

B1 being force time product needed for first mark

2

(c) (i) use of = 0.045 × 58 / 180 × 10−6

C1 use of 35 can gain first mark

or a = 58 / 180 = 3.2 × 105 (ignore power for first mark) 1.45 × 104 (N)

A1 2

(ii) (−)1.45 × 104 (N)

B1 numerically equal to c(i)

1

(iii) club head has inertia

C1 do not credit reference to friction

club head only slows slightly on impact

A1

club head still has kinetic energy / collision not elastic increase in internal energy / ߢheat’ / temperature of ball / club head

treat references to sound neutrally 2 max

[9]

Q12.

(a) (i) use of OR t 2 = 2s / g ✓

t = ✓ = 0.49 (0.4946 s) ✓ allow 0.5 do not allow 0.50

Some working required for full marks. Correct answer only gets 2

3

(ii) (s = vt ) = 8.5 × 0.4946 ✓ ecf ai = 4.2 m ✓ (4.20) ecf from ai

2

(b) (i)

t = or correct sub into equation above ✓

= = 8.2 × 10−2 (s) ✓ (0.0824) allow 0.08 but not 0.080 or 0.1

Allow alternative correct approaches

2

(ii) a = (v − u) / t OR correct substitution OR a = 103 ✓ ( = −8.5 ) / 8.24 × 10−2 = 103.2 )

(F = ma = ) 75 × (103.2) ✓ ecf from bi for incorrect acceleration due to arithmetic error only, not a physics error (e.g. do not allow a = 8.5. Use of g gets zero for the question.

= 7700 N ✓ (7741) ecf (see above) Or from loss of KE Some working required for full marks. Correct answer only gets 2

3 [10]

Q13. (a) ANY 2 from

• Slow moving neutrons or low (kinetic) energy neutrons

B1

• (They are in) thermal equilibrium with the moderator / Are in thermal equilibrium with other material (at a temperature of about 300 K)

B1

• Have energies of order of 0.025 eV

• Have (range of) KE similar to that of a gas at 300 K or room temperature

2

(b) (i) Use of mgh = ½ mv 2 by substitution or rearranges to make h the subject

PE for use of equation of motion (constant acceleration)

C1

0.086(1) (m) or 0.086(2) (m)

A1 2

(ii) Correct equation for conservation of momentum m1u1 (+ m2u2)= m1v1 + m2v2

or states momentum before = momentum after or pbefore= pafter

B1

(Correct clear Manipulation =) 0.065 (+ 0) = − 0.0325 + 0.0975 or −0.065 (+ 0) = 0.0325 − 0.0975 must see signs

Condone non−SI here: 65 (+0) = − 32.5 + 97.5

B1

States initial kinetic energy = final kinetic energy or States kinetic energy is conserved

Allow equivalent on RHS where masses are summed in one KE term

B1

(Correct clear Manipulation=) 0.04225 = 0.0105625 + 0.0316875

Or equivalent workings with numbers seen

and 0.04225 = 0.04225 / KE before = KE after

B1 4

(iii) (Percentage / fraction remaining after 1 collision =) ¼ = 25% seen

C1 OR % remaining = 100 × ½ m(1.32 − 0.652)/ ½ m1.32 or hockey ball = 0.0317 and initial ke = 0.04225 or their KE hb / 0.04225 or their KEhb / their KET

75(%) range 75 to 76

A1 2

(iv) Demonstrates: Slowing down / loss of KE of golf ball is like neutrons slowed down / Neutrons can lose KE by elastic collisions also

B1

Differs: Collisions in a reactor are not always / rarely head-on or KE loss is variable or Collisions are not always elastic or Ratio of mass of neutron to mass of nucleus is usually much smaller in a reactor

B1 2

(v) Water

B1 1

[13]

Q14. (a) 7.3(4) × 105 ✔

Numerical answer (in terms of powers of 10) must match unit prefixes where used Penalise rounding errors (733944.9541)

C kg–1 ✔ Do not allow use of solidus in unit: C / kg Condone a capital k or lower case c but not a capital g

2

(b) (1300 (eV) =) 2.08 × 10–16(J)

OR

2.1 × 10–16(J) ✔ 1

(c) Correct answer of 3.59 × 107 gains 3 marks (without working)

(Number of Xe ions per second) =

OR 1(.01) × 1019 seen ✔ Ecf from part (b)

(Mass of Xe ions per second)

= 2(.2) × 10–6 ✔ Ecf from part (b)

OR

(Total number of Xe ions) =

OR 3.6 × 1026 seen ✔ Ecf from part (b)

(total energy available)

3.6 × 1026 × (ans to (b)) OR 7.5(4) × 1010 ✔

Ecf from part (b)

✔ If both ‘methods’ attempted, restrict marks awarded to optimum method.

3

(d) Speed of He ions will be greater ✔

(Momentum depends on mass and speed, although) He (has higher speed) has (considerably) less mass, therefore less momentum (gained by He ion during the acceleration) ✔

He ion exerts less thrust (on spacecraft therefore xenon is better)

OR

Xenon ion exerts more thrust (on spacecraft therefore xenon is better) ✔

Must address these points Other points (e.g. He smaller so more can be stored) are neutral: no credit awarded Must be clear about which ion candidate is discussing Condone use of terms such as ‘heavier’ / ‘lighter’

3 [9]

Q15. (a) F∆t = ∆p (or stated in words)

∆p = 0.14 × 0.4 or 0.056 Accept use of F = ma with / in F∆t Accept 4.67 × 12 × 10-3 = 0.056

2

(b) 0.056 / 12 × 10-3 or 0.06 / 12 × 10-3 4.7 (4.67) (N)

Condone power of 10 error in t for C1 mark Allow 1 sf answer of 5 (N) for 0.06 (kg m s-1)

2 [4]

Q16. (a) no external forces (act on the system of particles)

[or forces between particles are internal forces] ✔ Allow “in a closed system”.

1

(b) (i) N V = (–) m v [or N V + m v = 0] ✔

(gives) V = (–) ✔ For 2nd mark, V must be the subject of the eqn.

2

(ii) ½ NV2 + ½ mv2 = E ✔

substitution for V gives + Ea = E ✔

from which + Ea = E and = E ✔

rearrangement gives ✔ The 4 marks are for • conservation of energy • substitution for V • separation of E and Eα, with v eliminated • rearrangement to give final result Allow ECF for incorrect V expression from (b)(i): for 1st and 2nd marks only (ie max 2).

4

(c) (i) nucleon number = 216 1

(ii) Eα = × 1.02 × 10–12 or = 1.00 × 10–12 (J) ✔

momentum of α = ✔ [or ½ × 4 × 1.66 × 10–27 × v2 = 1.00 × 10–12

gives momentum of α = 4 × 1.66 × 10–27 × ∴ momentum of α = ✔

= 1.2 (1.15) × 10–19 ✔ N s or kg m s–1 ✔ Allow ECF for wrong value of A from (c)(i). Alternative solution for first three marks: energy of nucleus = 0.0185 × 10–12 (J) ✔

momentum of nucleus = ✔

= 1.2 (1.15) × 10–19 ✔ Unit mark is independent.

4

(d) an (anti)neutrino is emitted

OR

two particles are emitted by unstable nucleus in β– decay

[or calculation must account for momentum of (anti)neutrino] ✔

[or momentum is shared between three particles] ✔ 1

[13]

Q17. (a) Max GPE of block = Mgh = 0.46 × 9.81 × 0.63 = 2.84 J ✓

The first mark is for working out the GPE of the block 1

Initial KE of block = ½ Mv2 = 2.84 J

Initial speed of block v2 = (2 × 2.84) / 0.46

v = 3.51 ms–1 ✓ The second mark is for working out the speed of the block initially

1

momentum lost by pellet = momentum gained by block

= Mv = 0.46 × 3.51 = 1.61 kg m s–1 ✓ The third mark is for working out the momentum of the block (and therefore pellet)

1

Speed of pellet = 1.58 / m = 1.58 / 8.8 × 10−3 = 180 ms–1 (183) ✓ The final mark is for the speed of the pellet

1

At each step the mark is for the method rather than the calculated answer Allow one consequential error in the final answer

(b) As pellet rebounds, change in momentum of pellet greater and therefore the change in momentum of the block is greater ✓

Ignore any discussion of air resistance 1

Initial speed of block is greater ✓ 1

(Mass stays the same)

Initial KE of block greater ✓ 1

Therefore height reached by steel block is greater than with wooden block ✓ 1

(c) Calculation of steel method will need to assume that collision is elastic so that change of momentum can be calculated ✓

1

This is unlikely due to deformation of bullet, production of sound etc. ✓ 1

And therefore steel method unlikely to produce accurate results. [10]

Q18. (a) arrow parallel to slope labelled (M+2m)gsin35 and label

parallel to slope labelled tension OR T ✔

Ignore arrows not parallel to ground e.g. weight Ignore friction W not acceptable for (M +2m)g

1

(b) T – Mgsin35 = Ma AND (M+2m)gsin35 – T = (M+2m)a ✔

add two equations

(M + 2m)gsin35 – Mg sin35 = Ma + (M +2m)a ✔

HENCE (a= mgsin35 / (M+m))

OR (M +2m)gsin35 – Mgsin35 ✔ (= (2M+2m)a)

a = 2mgsin35 /(2M +2m) ✔

HENCE (a= mgsin35 / (M+m))

1 1

(c) SECOND MARK CONDITIONAL ON FIRST mass / impulse / acceleration (of trollies) is the same ✔ momenta (trolley A and B) the same

SECOND MARK CONDITIONAL ON FIRST both have same speed / magnitude of velocity but different masses ✔ (hence) momentum of A is greater / momenta in opposite directions ✔

1 1

(d) ✔

(use of v2 =2as)

v = √(2 × 0.338 × 9.0) = 2.47 ✔

✔

OR

(use of s =1/2at2) 9 = ½ × 0.338 × t2 ✔ t = 7.3 s ✔

CE from acceleration calculation If used g for acceleration then no marks awarded

1 1 1

(e) number of journeys = (1800/(12 + 7.3) = 93 or 94 ✔

number of blocks = 2 × 93 = 186 or 2 × 94 = 188 ✔ Allow CE from 06.4 Allow between 93 to 94 Allow CE from incorrect number of journeys Allow 186 to 188

1 1

[10]

Q19. (a) m = 16 g = 0.016 kg r = 0.008 m

Use of V = 4 / 3 π r3 to give V = 4 / 3 π (0.008)3

= 2.1 × 10 − 6 m3✓ The first mark is for calculating the volume

1

Use of density = m / V to give density = 0.016 / 2.1 × 10−6✓ The second mark is for substituting into the density equation using the correct units

1

Density = 7.4 × 103 kg m−3 ✓ The final mark is for the answer.

1

(b) Use of v2 = u2 + 2as to give v2 = 2 (9.81) (1.27) ✓ (allow use of mgΔh = ½ mv2)

v2 = 25 (24.9) The first mark is for using the equation

1

v = 5.0 (m s-1) ✓ The second for the final answer

1

(c) Use of v2 = u2 + 2as to give 0 = u2 + 2 (-9.81) (0.85) ✓ The first mark is for using the equation

1

u2 = 17 (16.7)

u = 4.1 m s-1 ✓ The second for the final answer

1

(d) Change in momentum = mv + mu = 0.016 × 5 + 0.016 × 4.1✓ The first mark is for using the equation

1

= 0.15 (0.146) kg m s-1 ✓ The second for the final answer

1

(e) Use of Force = change in momentum / time taken

= 0.15 / 40 × 10-3✓ The first mark is for using the equation

1

= 3.6 N ✓ The second for the final answer

1

(f) Impact time can be increased if the plinth material is not stiff✓ Alternative A softer plinth would decrease the change in momentum of the ball (or reduce the height of rebound) ✓

1

Increased impact time would reduce the force of the impact. ✓ Smaller change in momentum would reduce the force of impact ✓

1 [13]

Q20. C

[1]

Q21. B

[1]

Q22. A

[1]

Q23. C

[1]

Q24.

B [1]

Q25. D

[1]

Q26. A

[1]

Q27. D

[1]

Q28. D

[1]

Q29. C

[1]

Q30. D

[1]

Q31. D

[1]

Q32. B

[1]

Q33. B

[1]

Q34. B

[1]

Q35. D

[1]