Mathematics 9709/42 February/March 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A crate of mass 200 kg is being pulled at constant speed along horizontal ground by a horizontal rope attached to a winch. The winch is working at a constant rate of 4.5 kW and there is a constant resistance to the motion of the crate of magnitude 600 N.
(a) Find the time that it takes for the crate to move a distance of 15 m. [2]
The rope breaks after the crate has moved 15 m.
(b) Find the time taken, after the rope breaks, for the crate to come to rest. [3]
Mark scheme
1(a) 600 × 15 M1 Use of power = ∆W / ∆t to get an equation in t .
4500 = May see 600v = 4500 v = 7.5 followed by 7.5t = 15 .
t
t =2s A1
2
1(b) ±600 = 200a [ a = ±3] *M1 Use of Newton’s second law; 2 terms only.
15 DM1 Use of constant acceleration to set up an equation that would
0= + ( their − 3) t lead to a positive t , e.g. v = u + at with their t = 2 and their
their 2
negative a (and possibly their 7.5 from (a)).
t = 2.5s A1
3
A particle P is projected vertically upwards from horizontal ground with speed 15 m s−1 .
(a) Find the speed of P when it is 10 m above the ground. [2]
At the same instant that P is projected, a second particle Q is dropped from a height of 18 m above the ground in the same vertical line as P.
(b) Find the height above the ground at which the two particles collide. [3]
Mark scheme
2(a) Use constant acceleration in an attempt to find v or v 2 M1 e.g. v 2 = u 2 + 2as with a = ± g .
[v2 = 152 – 2g × 10]
Speed = 5 m s-1 A1
2
( sP = ) ± 15t −
2(b) 1 2 *B1
( s =) ± 12 gt
1
gt , Q
2
Use of s = ut + at 2 for either.
2 2
Allow if a not substituted, need both expressions with opposite
sign of t 2 term and the same a .
Use sP + sQ = 18 and solve for t DM1 Allow sP + sQ = ±18 .
Must have s P and sQ of the correct form.
So height = 10.8 m A1
Alternative method for Question 2(b): Using relative velocity
±15t *B1 Use of relative velocity (no acceleration).
Use 15t = 18 and solve for t DM1 Allow 15t = ±18 .
So height = 10.8 m A1 Not from t = −1.2 made positive without justification.
3
A particle moves in a straight line starting from rest from a point O. The acceleration of the particle
1 at time t s after leaving O is a m s−2 , where a = 4t 2 .
(a) Find the speed of the particle when t = 9. [2]
(b) Find the time after leaving O at which the speed (in metres per second) and the distance travelled
(in metres) are numerically equal. [3]
Mark scheme
3(a) Attempt to integrate M1 Increasing power by 1 and a change in coefficient in at least one
4
3
8
3
term; may be unsimplified.
( v = ) t 2 = t 2 ( + c ) v = at M0.
1.5 3
Substitute t = 9 to get speed = 72 m s–1 A1 Or use limits t = 0 and t = 9 .
2
3(b) Attempt at integration of their v *M1 Increasing power by 1 and a change in coefficient in at least one
8 5 term; may be unsimplified.
( s = ) = 3 t 2 = 16 t 2 ( +c ') s = vt M0
2.5 15 Their v , which has come from integration in part (a).
Equate their v and their s and attempt to solve for t DM1 Their v must have come from integration.
16 5 8 3 16 8 Allow if their c from (a) is not 0.
t 2 = t 2 t − = 0
15 3 15 3
5 A1 5
time = s Must discard t = 0 and t = − .
2 2
3
A toy railway locomotive of mass 0.8 kg is towing a truck of mass 0.4 kg on a straight horizontal track at a constant speed of 2 m s−1 . There is a constant resistance force of magnitude 0.2 N on the locomotive, but no resistance force on the truck. There is a light rigid horizontal coupling connecting the locomotive and the truck.
(a) State the tension in the coupling. [1]
(b) Find the power produced by the locomotive’s engine. [1]
The power produced by the locomotive’s engine is now changed to 1.2 W.
(c) Find the magnitude of the tension in the coupling at the instant that the locomotive begins to accelerate. [5]
Mark scheme
4(a) Tension = 0 N B1 May be implied.
1
4(b) Power [ = 0.2 × 2] = 0.4 W B1 Use of power = Fv .
Allow without units.
1
4(c) Driving force = 1.2/2 [= 0.6 N] B1
Use of Newton’s second law for locomotive or truck or system M1 Correct number of relevant terms.
For locomotive: DF – 0.2 –T = 0.8a A1 For any two correct.
For truck: T = 0.4a
For system: DF –0.2 =1.2a
For attempt to solve for T M1 From equations with correct number of relevant terms.
Using their dimensionally correct DF.
1
May see a = .
3
2 A1 Allow awrt 0.133 .
T= N
15
5
C
500 N
D 100 kg
45Å 45Å
A B
The diagram shows a block D of mass 100 kg supported by two sloping struts AD and BD, each attached at an angle of 45Å to fixed points A and B respectively on a horizontal floor. The block is also held in place by a vertical rope CD attached to a fixed point C on a horizontal ceiling. The tension in the rope CD is 500 N and the block rests in equilibrium.
(a) Find the magnitude of the force in each of the struts AD and BD. [3]
A horizontal force of magnitude F N is applied to the block in a direction parallel to AB.
(b) Find the value of F for which the magnitude of the force in the strut AD is zero. [3]
Mark scheme
5(a) Attempt to resolve vertically M1 4 terms; allow with TA and TB ; allow sign errors; allow g
missing.
500 + T cos 45 + T cos 45 − 100 g = 0 A1 Must have TA = TB = T .
Allow if 500 − 2T cos 45 − 100 g = 0 .
OR 500 + TA cos 45 + TB cos 45 − 100 g = 0 AND Allow 500 − TA cos 45 − TB cos 45 − 100 g = 0 AND
TA ( sin 45 ) = TB ( sin 45 ) TA ( sin 45 ) = TB ( sin 45 ) .
T = 354 N A1 500
Allow 250 2 , .
2
Allow if 500 − 2T cos 45 − 100 g = 0 to obtain
T =−354 and then state magnitude is 354.
If TA and TB are different values then A0.
Alternative Method 1 for Question 5(a): Resolving perpendicular to a strut
Resolve perpendicular to TA or TB M1 3 terms; allow sign errors; allow g missing.
TA ( or TB ) + 500cos 45 = 100 g cos 45 A1 Allow TA ( or TB ) + 100 g cos 45 = 500cos 45 .
TA = TB = 354 A1 500
Allow 250 2 , .
2
5(a) Alternative Method 2 for Question 5(a): Using triangle of forces
Attempt Pythagoras on a right-angled triangle of forces or use of M1 4 terms; allow with TA and TB ; allow sign errors; allow g
trigonometry missing.
TA2 + TB 2 = (100 g − 500 )
2
OR sin45or cos 45 = 100 g − 500 or 100 g − 500
TA TB
T 2 + T 2 = (100 g − 500) A1
2
OR TA2 + TB 2 = (100g − 500) AND TA ( sin 45 ) = TB ( sin 45 )
2
TA ( or TB ) TA ( or TB )
Allow sin 45 = OR cos 45 = .
TA ( or TB ) TA ( or TB ) 500 − 100 g 500 − 100 g
OR sin 45 = OR cos 45 =
100 g − 500 100 g − 500
TA = TB = 354 A1 500
Allow 250 2 , .
2
Alternative Method 3 for Question 5(a): Using Lami’s Theorem
Attempt at Lami M1 Allow with TA and TB ; allow sign errors; allow g missing.
100 g − 500 TA ( or TB ) A1 500 − 100 g TA ( or TB )
= Allow = .
sin 90 sin135 sin 90 sin135
100 g − 500 TA ( or TB )
Allow = .
sin 270 sin 45
TA = TB = 354 A1 500
Allow 250 2 , .
2
3
5(b) Attempt to resolve vertically and horizontally M1 3 terms vertically and 2 terms horizontally; allow sign errors;
allow g missing.
Must have TA = 0 .
TB cos 45 + 500 − 100 g = 0 and A1 Allow −TB cos 45 + 500 − 100 g = 0 and
F − TB sin 45 = 0 F + TB sin 45 = 0
OR TB cos 45 + 500 − 100 g = 0 and
F + TB sin 45 = 0
OR −TB cos 45 + 500 − 100 g = 0 and
F − TB sin 45 = 0 .
For both equations correct.
F = 500 A1 awrt 500 to 3sf.
Alternative Method 1 for Question 5(b): Resolving perpendicular to TB
Attempt to resolve perpendicular to TB M1 3 terms; allow sign errors; allow g missing.
Must have TA = 0 .
F cos 45 + 500cos 45 = 100 g cos 45 A1 Allow −F cos 45 + 500cos 45 = 100g cos 45 .
F = 500 A1 awrt 500 to 3sf.
5(b) Alternative Method 2 for Question 5(b): Using Lami’s Theorem
Attempt at Lami M1 Allow sign errors.
100 g − 500 F TB A1 500 − 100 g F 100 g − 500 F
= = Allow = or = or
sin135 sin135 sin 90 sin135 sin135 sin 45 sin 45
100 g − 500 F 100 g − 500 F
= or = .
sin 45 sin 225 sin 225 sin 45
F = 500 A1 awrt 500 to 3sf.
Alternative Method 3 for Question 5(b): Using triangle of forces
Attempt use of trigonometry on right angled triangle M1 Allow sign errors; allow g missing.
F A1 F
tan45= Allow tan45=
100 g − 500 500 − 100 g
F = 500 A1 awrt 500 to 3sf.
3
TN
20Å
B
30Å
FN
A block B, of mass 2 kg, lies on a rough inclined plane sloping at 30Å to the horizontal. A light rope, inclined at an angle of 20Å above a line of greatest slope, is attached to B. The tension in the rope is T N. There is a friction force of F N acting on B (see diagram). The coefficient of friction between
B and the plane is -.
(a) It is given that F = 5 and that the acceleration of B up the plane is 1.2 m s−2 .
(i) Find the value of T . [3]
(ii) Find the value of -. [3]
(b) It is given instead that - = 0.8 and T = 15.
Determine whether B will move up the plane. [3]
Mark scheme
6(a)(i) Attempt to resolve parallel to the plane M1 4 terms; allow sin/cos mix; allow sign errors; allow g missing.
Tcos20 − 5 − 2 gsin30 = 2 ×1.2 A1 Correct equation.
T = 18.5 A1 awrt 18.5 .
3
6(a)(ii) Attempt resolve perpendicular to the plane M1 3 terms; allow sin/cos mix: allow sign errors; allow with T or
R = 2 g cos30 − Tsin20 their T ; allow g missing.
Use of 5 = μ R to get an equation in μ only M1 Where R is a two term expression with a component of 2g and
5 = μ × ( 2g cos30 − Tsin20) a component of their T ; allow g missing.
μ = 0.455 A1 5
awrt 0.455; allow 0.46 or 0.45; do not allow .
11
3
6(b) Max F = 0.8 × ( 2 g cos30 − 15sin20 ) *B1
[ = 0.8 × 12.1902 = 9.7521]
Net force up the plane = 15cos20 − 2 gsin30 [ = 4.0953] *B1 15cos20 − 2 gsin30 − 0.8 × ( 2 g cos30 − 15sin20 )
OR
[ = 2a] −5.6567[= 2a] a = −2.8283..
If max F incorrect and use F = ma then allow B1 for
15cos20 − 2 gsin30 − their max F .
[State 4.0953< 9.7521 ,] hence the block does not move DB1 Must have correct values (to at least 1 sf) to compare for this
[up the plane] mark. No incorrect statement seen.
Alternative Method 1 for Question 6(b)
Max force down plane = 0.8 × ( 2 g cos30 − 15sin20 ) + 2 gsin30 *B1
[ = 0.8 × 12.1902 + 10 = 19.7521]
Force up plane = 15cos20 [ = 14.0953] *B1 i.e. using it to compare with their max force down the plane.
[State 1 4.0953< 19.7521 ,] hence the block does not move DB1 Must have correct values (to at least 1 sf) to compare for this
[up the plane] mark. No incorrect statement seen.
6(b) Alternative Method 2 for Question 6(b)
F = 15cos20 − 2 gsin30 [ = 4.9053] *B1
Or R = 2 g cos30 − 15sin20 [ = 12.1902]
15cos20 − 2 gsin30 *B1
Get μ = [ = 0.3359]
2 g cos30 − 15sin20
[State 0.3359< 0.8 ,] hence the block does not move [up the DB1 Must have correct value of μ (to at least 1 sf) to compare for
plane] this mark. No incorrect statement seen.
3
O
A
E
1.8 m
F
1Å
B 7.0 m C
The diagram shows a smooth track which lies in a vertical plane. The section AB is a quarter circle of radius 1.8 m with centre O. The section BC is a horizontal straight line of length 7.0 m and OB is perpendicular to BC. The section CFE is a straight line inclined at an angle of 1Å above the horizontal.
A particle P of mass 0.5 kg is released from rest at A. Particle P collides with a particle Q of mass
0.1 kg which is at rest at B. Immediately after the collision, the speed of P is 4 m s−1 in the direction
BC. You should assume that P is moving horizontally when it collides with Q.
(a) Show that the speed of Q immediately after the collision is 10 m s−1 . [4]
When Q reaches C, it collides with a particle R of mass 0.4 kg which is at rest at C. The two particles coalesce. The combined particle comes instantaneously to rest at F. You should assume that there is no instantaneous change in speed as the combined particle leaves C, nor when it passes through C again as it returns down the slope.
(b) Given that the distance CF is 0.4 m, find the value of 1. [4]
[Question 7 continues on the next page.]
(c) Find the distance from B at which P collides with the combined particle. [5]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) Attempt to use conservation of energy M1 2 terms, dimensionally correct.
1 1 Do not allow from use of constant acceleration.
2 × 0.5v = 0.5 g × 1.8 or 2 × mv = mg × 1.8
2 2
v=6 A1 Do not allow from use of constant acceleration.
Attempt at conservation of momentum M1 3 terms; allow sign errors; allow their v = 6 or just v ; allow if
0.5 × 6 ( +0) = 0.5 × 4 + 0.1w using mgv (consistently in all terms).
Speed of Q ( = w ) =10 m s-1 A1 AG
Do not allow from use of constant acceleration.
Do not allow if using mgv .
Use of constant acceleration gets M0 A0 M1 A0 maximum.
4 SC Assuming elastic collision
1 1
M1A1 0.5 g ×1.8 = × 0.1w2 + × 0.5 × 42
2 2
M1 For attempt at conservation of energy, 3 terms; allow sign
errors.
B1 Speed of Q ( = w ) =10 m s-1
7(b) Attempt at conservation of momentum *M1 3 terms, allow sign errors, allow if using mgv .
0.1×10 = ( 0.1 + 0.4) × z ( z = 2)
Attempt to use conservation of energy *DM1 Dependent on previous M mark.
1 4 terms, dimensionally correct.
2 × ( 0.1 + 0.4 ) × ( their 2 ) = ( 0.1 + 0.4 ) gh ( h = 0.2 )
2
Do not allow from use of constant acceleration.
their 2 ≠ 10 .
Use trigonometry to get an equation in θ and solve for θ DM1 Dependent on previous 2 M marks.
−1 their 0.2
Using their h and 0.4 .
θ = sin 0.4 Allow sin/cos mix.
θ = 30 A1 Do not allow if using mgv .
Alternative method for Question 7(b): Using constant acceleration
Attempt at conservation of momentum *M1 2 terms, allow sign errors, allow if using mgv .
0.1×10 = 0.5 × z ( z = 2)
Attempt at use of constant acceleration *DM1 Dependent on previous M mark.
02 = ( their 2 )2 ± 2 × a × 0.4 ( a = 5 ) Uses constant acceleration with u = their 2 and s = 0.4 to get an
equation in a ; their 2 ≠ 10 .
Use N2L to get an equation in θ leading to a positive value of θ DM1 Dependent on previous 2 M marks.
and solve for θ Using their a ; May have m for 0.5 .
( 0.5) their a = ( 0.5) g sin θ Allow sin/cos mix.
θ = 30 A1 Do not allow if using mgv .
4
7(c) Q takes 0.7 s to travel from B to C B1
0.4 =
( their 2 ) + 0 t t = 0.4 B1FT SOI
0.8
2 FT their 2 from (b), t = .
their 2
u+v
For use of s = t to get a time up the slope.
2
Allow for total time on slope from
1
0 = ( their 2 ) t − ( their a ) t 2 t = 0.8 .
2
Distance between P moved is ( 0.7 + 0.8 ) × 4 ( = 6 ) B1 Allow 1 m from point C.
Set up equation in t using 4t , ( their 2 ) t and their 6 and solve for M1 Must have considered all parts of motion to find times from
t relevant equations.
4t + (their 2) t = (their 1) OR (their 6) + 4t + (their 2) t = 7
2 A1
Distance from B = 6 m
3
7(c) Alternative method for last 3 marks of Question 7(c)
b 7−b B1 Where b is distance from B
[Time for P =] and [Time for QR =]
4 2
7−c c OR Where c is distance from C.
OR [Time for P =] and [Time for QR =]
4 2
Attempt to form an equation from use of total time and solve for b M1 Where b is distance from B
(or c ) OR Where c is distance from C.
Must have considered all parts of motion to find times from
7−b b 2 relevant equations.
2 + 0.7 + 0.4 + 0.4 = 4 b = 6 3
c 7−c 1
OR + 0.7 + 0.4 + 0.4 = c =
2 4 3
2 A1
Distance from B = 6 m
3
5
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