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A-Level · Mechanics

M2 Mechanics: Forces and Newton's Laws

EDEXCEL 9MA0 · Calculator allowed · about 155 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A box of mass 15 kg is pulled in a straight line along rough horizontal ground by a light rope which is horizontal, attached to the box with tension 45 N. The resistance to motion from the ground is modelled as a constant force of 15 N. The box is modelled as a particle. Take g = 9.8 m/s^2.

(a)State the four forces acting on the box while it is moving.(2)

(b)Use Newton's second law to find the acceleration of the box.(3)

(c)Find the magnitude of the normal reaction between the box and the ground.(2)

(d)Given that the box starts from rest, find the distance it travels in the first 6 seconds of motion.(3)

(Total for Question 1 is 10 marks)

Priya, of mass 70 kg, stands on the floor of a lift. Take g = 9.8 m/s² and model Priya as a particle. In each case, find the normal reaction force R exerted on Priya by the lift floor (her apparent weight).

(a)The lift accelerates upwards at 1.5 m/s^2. Find R.(3)

(b)The lift instead accelerates downwards at 1.5 m/s^2. Find R.(3)

(c)The lift moves at a constant velocity of 2 m/s upwards. Find R.(2)

(d)Explain, with reference to your answers above, why Priya feels heavier when the lift accelerates upwards and lighter when it accelerates downwards.(2)

(Total for Question 2 is 10 marks)

Two blocks A (mass 5 kg) and B (mass 3 kg) are connected by a light inextensible string. Block A is pulled by a horizontal force of magnitude 32 N, with the string joining A to B, which is behind A, so that A and B move together in a straight line with the same acceleration. Take g = 9.8 m/s^2.

(a)Given that the surface is smooth, find the acceleration of the system.(3)

(b)Find the tension in the string.(3)

(c)The surface beneath B is now rough, producing a constant resistance of 8 N to B's motion; the surface beneath A remains smooth. Find the new acceleration of the system and the new tension in the string.(4)

(Total for Question 3 is 10 marks)

Two particles, of mass M kg and m kg (M > m), are connected by a light inextensible string which passes over a smooth fixed pulley. The particles hang freely on either side of the pulley and are released from rest. Take g = 9.8 m/s^2.

(a)By writing separate equations of motion for each particle, show that the acceleration of the system is a = g(M - m)/(M + m).(4)

(b)Given that M = 5 kg and m = 3 kg, find the acceleration of the system and the tension in the string.(4)

(c)State two modelling assumptions used in this problem, and explain the effect on the acceleration if the pulley were not smooth.(2)

(Total for Question 4 is 10 marks)

A block of mass 8 kg rests in equilibrium on a smooth plane inclined at angle θ to the horizontal, held by a light inextensible string parallel to the line of greatest slope, attached to a fixed point higher up the plane. Take g = 9.8 m/s^2.

(a)Given that θ = 20 degrees, find the tension in the string.(3)

(b)Find the normal reaction between the block and the plane.(3)

(c)The plane is replaced by one inclined at 35 degrees (all else unchanged). Find the new tension in the string.(2)

(d)Explain why the normal reaction decreases as the angle of the incline increases.(2)

(Total for Question 5 is 10 marks)

A crate of mass 20 kg rests on a rough plane inclined at 15 degrees to the horizontal. The coefficient of friction between the crate and the plane is 0.35. The crate is modelled as a particle. Take g = 9.8 m/s^2.

(a)Show that the crate remains in equilibrium on the plane.(3)

(b)Find the magnitude of the friction force acting on the crate while it is at rest.(2)

(c)A force P is now applied to the crate, parallel to and directed up the line of greatest slope. Find the minimum value of P required to move the crate up the plane.(4)

(d)State one assumption made about the crate in this model.(1)

(Total for Question 6 is 10 marks)

Particle A, of mass 6 kg, lies on a rough plane inclined at angle θ to the horizontal. A is connected by a light inextensible string, passing over a smooth light pulley fixed at the top of the plane, to particle B, of mass 4 kg, which hangs freely below the pulley. The coefficient of friction between A and the plane is 0.25. The system is released from rest. Take g = 9.8 m/s^2.

(a)Given that θ = 30 degrees, show that the system remains in equilibrium.(4)

(b)The plane is instead inclined at 60 degrees (all else unchanged). Show that the system does not remain in equilibrium, and find the resulting acceleration of the system.(6)

(Total for Question 7 is 10 marks)

A particle P of mass 2 kg moves on a smooth horizontal plane under the action of two constant forces, F1 = (3i + 5j) N and F2 = (-i + 2j) N, where i and j are perpendicular unit vectors in the plane. The weight of P and the normal reaction from the plane are not considered further, as they act perpendicular to the plane of motion.

(a)Find the resultant force acting on P, giving your answer in the form (π + qj) N.(2)

(b)Find the acceleration of P, in the form (π + qj) m/s^2.(2)

(c)Find the magnitude of the acceleration, giving your answer to 3 significant figures.(2)

(d)Given that P starts at rest, find the speed of P after 4 seconds, giving your answer to 3 significant figures.(3)

(e)Find, to the nearest degree, the angle the direction of motion after 4 seconds makes with the vector i.(2)

(Total for Question 8 is 11 marks)

A lorry of mass 1200 kg tows a trailer of mass 400 kg along a straight horizontal road, connected by a light rigid horizontal tow-bar. The lorry's engine produces a driving force of 3000 N. The resistances to motion are 500 N on the lorry and 300 N on the trailer. Both vehicles are modelled as particles, and the tow-bar as a light rod. Take g = 9.8 m/s² (not required numerically in this question).

(a)Find the acceleration of the lorry and trailer.(3)

(b)Find the force in the tow-bar, stating whether it is a tension or a thrust.(3)

(c)The engine is switched off while both vehicles are moving, so the driving force becomes zero; the resistances are unchanged. Find the new deceleration of the system and the new force in the tow-bar, stating whether it is a tension or a thrust.(4)

(d)State an assumption made about the tow-bar in this model, and explain why it allows the lorry and trailer to have the same acceleration.(2)

(Total for Question 9 is 12 marks)

A particle P of weight 45 N is held in equilibrium by two light inextensible strings. One string is inclined at 50 degrees to the horizontal and is attached to a fixed point above and to one side of P; the other string is horizontal, attached to a fixed point on a wall.

(a)Find the tension T1 in the inclined string.(3)

(b)Find the tension T2 in the horizontal string.(3)

(c)The horizontal string is cut. Find the angle the remaining string now makes with the horizontal, and the new tension in it, for P to hang in equilibrium.(3)

(Total for Question 10 is 9 marks)

A block of mass 10 kg rests on a rough plane inclined at 20 degrees to the horizontal. The coefficient of friction between the block and the plane is 0.4. A horizontal force of magnitude P newtons is applied to the block, directed into the slope, so that it has a component acting up the line of greatest slope. Take g = 9.8 m/s^2.

(a)Given that P = 0, show that the block remains in equilibrium.(3)

(b)Find, to 3 significant figures, the minimum value of P required to move the block up the plane.(6)

(Total for Question 11 is 9 marks)

Particle A, of mass 2 kg, lies on a rough horizontal table. Particle A is connected by a light inextensible string, passing over a smooth light pulley fixed at the edge of the table, to particle B, of mass 3 kg, which hangs freely 1.2 m above the floor. The coefficient of friction between A and the table is 0.3. The system is released from rest with the string taut. Take g = 9.8 m/s^2.

(a)Find the acceleration of the system and the tension in the string while B is falling.(6)

(b)Find the speed of B just before it reaches the floor.(3)

(c)When B lands, it does not rebound and the string becomes slack. Particle A continues moving across the table, decelerating due to friction alone. Find the additional distance A travels before it comes to rest.(5)

(d)State one modelling assumption made about the string, and explain what would happen to your answer to part (c) if the string had mass.(2)

(Total for Question 12 is 16 marks)

Mark scheme · M2 Mechanics: Forces and Newton's Laws

Question 1

(a) B1 weight (mg, vertically down) and normal reaction (R, vertically up) identified
(a) B1 tension (T = 45 N, horizontal, direction of motion) and resistance (15 N, opposing motion) identified
(a) Answer: Weight, normal reaction, tension (45 N), resistance (15 N)
(b) M1 resolving horizontally: 45 - 15 = 15a (oe)
(b) A1 net force = 30 N found
(b) A1 a = 2 m/s² cao
(b) Answer: a = 2 m/s²
(c) M1 resolving vertically: R = mg (since rope is horizontal)
(c) A1 R = 147 N cao
(c) Answer: R = 147 N
(d) M1 correct suvat selected, s = ut + 1/2 a t² with u = 0, a = 2, t = 6
(d) A1 correct substitution s = 0.5 x 2 x 36 (oe)
(d) A1 s = 36 m cao
(d) Answer: 36 m

Question 2

(a) M1 equation of motion R - mg = ma with a = +1.5 (upward taken as positive)
(a) A1 R = 70(9.8 + 1.5) (oe)
(a) A1 R = 791 N cao
(a) Answer: R = 791 N
(b) M1 equation of motion mg - R = ma with a = 1.5 downward, or R = m(g - a)
(b) A1 R = 70(9.8 - 1.5) (oe)
(b) A1 R = 581 N cao
(b) Answer: R = 581 N
(c) B1 recognising a = 0 so R = mg
(c) B1 R = 686 N cao
(c) Answer: R = 686 N
(d) B1 R > mg (791 > 686) when accelerating upwards, and R is what Priya feels supporting her, so she feels heavier
(d) B1 R < mg (581 < 686) when accelerating downwards, so she feels lighter
(d) Answer: R is greater than her true weight when accelerating up, and less than her true weight when accelerating down; apparent weight equals R.

Question 3

(a) M1 whole system equation: 32 = (5 + 3)a (oe)
(a) A1 8a = 32 (oe)
(a) A1 a = 4 m/s² cao
(a) Answer: a = 4 m/s²
(b) M1 equation of motion for B alone: T = mB a (oe)
(b) A1 T = 3 x 4 (oe)
(b) A1 T = 12 N cao
(b) Answer: T = 12 N
(c) M1 whole system equation: 32 - 8 = (5+3)a (oe)
(c) A1 a = 3 m/s² cao
(c) M1 equation of motion for B: T - 8 = 3a
(c) A1 T = 17 N cao
(c) Answer: a = 3 m/s², T = 17 N

Question 4

(a) M1 equation of motion for M: Mg - T = Ma
(a) M1 equation of motion for m: T - mg = ma
(a) dM1 eliminate T by adding the two equations: Mg - mg = (M+m)a
(a) A1 cso: a = g(M-m)/(M+m) correctly rearranged and shown
(a) Answer: a = g(M - m)/(M + m) (given, shown)
(b) M1 substitute M=5, m=3 into a = g(M-m)/(M+m)
(b) A1 a = 2.45 m/s² (awrt 2.45)
(b) M1 substitute into T = M(g-a) or T = m(g+a)
(b) A1 T = 36.75 N (awrt 36.8)
(b) Answer: a = 2.45 m/s², T = 36.75 N
(c) B1 assumptions stated: string is light and inextensible; pulley is smooth and light (any two)
(c) B1 if the pulley were not smooth, friction at the pulley would resist the string's motion over it, reducing the net driving force and so the acceleration would be smaller than 2.45 m/s²
(c) Answer: Assumptions: light inextensible string; smooth, light pulley. A rough pulley reduces the acceleration.

Question 5

(a) M1 resolving parallel to the plane: T = mg sin(θ)
(a) A1 T = 8(9.8) sin20 (oe)
(a) A1 T = 26.8 N (awrt)
(a) Answer: T = 26.8 N
(b) M1 resolving perpendicular to the plane: R = mg cos(θ)
(b) A1 R = 78.4 cos20 (oe)
(b) A1 R = 73.7 N (awrt)
(b) Answer: R = 73.7 N
(c) M1 T = 78.4 sin35
(c) A1 T = 45.0 N (awrt)
(c) Answer: T = 45.0 N
(d) B1 R = mg cos(θ), and cos(θ) decreases as θ increases for 0 < θ < 90 degrees
(d) B1 so the component of weight perpendicular to the plane decreases as the plane steepens, reducing R
(d) Answer: R = mg cos(θ) decreases because cos(θ) decreases as θ increases.

Question 6

(a) M1 compute component of weight down the plane: mg sin15 = 50.7 N (awrt)
(a) M1 compute maximum available friction: μ mg cos15 = 66.3 N (awrt)
(a) A1 cso: since 50.7 < 66.3, the required friction is within the limiting value, so the crate remains in equilibrium
(a) Answer: Remains in equilibrium (50.7 N < 66.3 N)
(b) B1 recognising friction is not limiting, so F = mg sin15 (not μ R)
(b) B1 F = 50.7 N (awrt), acting up the plane
(b) Answer: F = 50.7 N up the plane
(c) M1 at limiting equilibrium (about to move up), friction acts down the plane at its maximum: F = μ mg cos15
(c) A1 equation: P = mg sin15 + μ mg cos15 (oe)
(c) M1 substitute values: P = 50.73 + 66.26
(c) A1 P = 117 N (awrt)
(c) Answer: P = 117 N
(d) B1 the crate is modelled as a particle (a single point with no size/shape), so rotational effects are ignored
(d) Answer: The crate is modelled as a particle.

Question 7

(a) M1 find R = mA g cos30 = 50.9 N (awrt) and maximum friction Fmax = μ R = 12.7 N (awrt)
(a) M1 compare tension required to hold B (T = mBg = 39.2 N) with weight component of A along plane (mA g sin30 = 29.4 N); required friction = |39.2 - 29.4| = 9.8 N
(a) A1 cso: since 9.8 N < 12.7 N (Fmax), the required friction is available, so the system remains in equilibrium
(a) Answer: Remains in equilibrium (required friction 9.8 N < max friction 12.7 N)
(b) M1 find new R = mA g cos60 = 29.4 N and Fmax = μ R = 7.35 N
(b) M1 compare mA g sin60 (50.9 N, awrt) with mBg (39.2 N): difference = 11.7 N (awrt)
(b) A1 cso: since 11.7 N > 7.35 N, friction cannot hold the system, so A slides down the plane and B rises; not in equilibrium
(b) M1 equation of motion for whole system: mA g sin60 - mBg - Fmax = (mA+mB)a
(b) A1 correct substitution: 50.92 - 39.2 - 7.35 = 10a
(b) A1 a = 0.44 m/s² (awrt)
(b) Answer: Not in equilibrium; a = 0.44 m/s² (A slides down, B rises)

Question 8

(a) M1 add components: (3-1)i + (5+2)j
(a) A1 resultant = (2i + 7j) N cao
(a) Answer: (2i + 7j) N
(b) M1 a = F/m = (2i+7j)/2
(b) A1 a = (1i + 3.5j) m/s² cao
(b) Answer: (i + 3.5j) m/s²
(c) M1 |a| = √1² + 3.5²
(c) A1 |a| = 3.64 m/s² (awrt)
(c) Answer: 3.64 m/s²
(d) M1 v = u + at with u = 0, t = 4, using the vector acceleration
(d) A1 v = (4i + 14j) m/s
(d) A1 speed = |v| = 14.6 m/s (awrt)
(d) Answer: 14.6 m/s
(e) M1 tan(angle) = 14/4 (oe, using components of v)
(e) A1 angle = 74 degrees (awrt, nearest degree)
(e) Answer: 74 degrees

Question 9

(a) M1 whole system equation: 3000 - 500 - 300 = (1200+400)a
(a) A1 2200 = 1600a (oe)
(a) A1 a = 1.375 m/s² cao
(a) Answer: a = 1.375 m/s²
(b) M1 equation of motion for trailer alone: T - 300 = 400a
(b) A1 T = 300 + 400(1.375)
(b) A1 T = 850 N cao, a tension (bar pulls trailer forward)
(b) Answer: T = 850 N (tension)
(c) M1 whole system equation with zero driving force: -800 = 1600a
(c) A1 deceleration = 0.5 m/s²
(c) M1 equation of motion for trailer: T - 300 = 400(-0.5)
(c) A1 T = 100 N cao, still a tension
(c) Answer: Deceleration = 0.5 m/s²; T = 100 N (tension)
(d) B1 the tow-bar is modelled as light and rigid (inextensible)
(d) B1 because it is rigid and inextensible, the distance between the lorry and trailer cannot change, so their velocities and accelerations must be equal at all times
(d) Answer: The tow-bar is light and rigid; this fixes the separation of the two vehicles so they share the same acceleration.

Question 10

(a) M1 resolving vertically: T1 sin50 = 45
(a) A1 T1 = 45/sin50 (oe)
(a) A1 T1 = 58.7 N (awrt)
(a) Answer: T1 = 58.7 N
(b) M1 resolving horizontally: T2 = T1 cos50 (oe, or T2 = 45/tan50)
(b) A1 T2 = 58.74 cos50 (oe)
(b) A1 T2 = 37.8 N (awrt)
(b) Answer: T2 = 37.8 N
(c) B1 with only one string and the weight acting, the string must be vertical (90 degrees to horizontal) for horizontal equilibrium
(c) M1 resolving vertically: T = 45 (weight only supported by the string)
(c) A1 angle = 90 degrees, T = 45 N
(c) Answer: The string becomes vertical (90 degrees); T = 45 N

Question 11

(a) M1 compare tan(θ) with μ, or equivalently compute mg sin20 and μ mg cos20
(a) A1 tan20 = 0.364 (awrt), or mg sin20 = 33.5 N and μ mg cos20 = 36.8 N (awrt)
(a) A1 cso: since tan20 (0.364) < μ (0.4) [equivalently 33.5 N < 36.8 N], the block remains in equilibrium
(a) Answer: Remains in equilibrium (tan20 = 0.364 < 0.4)
(b) M1 resolving perpendicular to the plane: R = mg cos20 + P sin20
(b) A1 correct expression for R in terms of P
(b) M1 resolving along the plane at limiting equilibrium (motion up impending, friction acting down the plane): P cos20 = mg sin20 + μ R
(b) A1 correct equation with R substituted: P cos20 = mg sin20 + μ(mg cos20 + P sin20)
(b) M1 rearrange to make P the subject: P(cos20 - μ sin20) = mg(sin20 + μ cos20)
(b) A1 P = 87.6 N (awrt)
(b) Answer: P = 87.6 N

Question 12

(a) M1 equation of motion for A: T - F = mA a, with friction F = μ mA g
(a) M1 equation of motion for B: mB g - T = mB a
(a) A1 F = 0.3(2)(9.8) = 5.88 N correctly found
(a) dM1 eliminate T by combining the two equations to solve for a
(a) A1 a = 4.70 m/s² (awrt)
(a) A1 T = 15.3 N (awrt)
(a) Answer: a = 4.70 m/s², T = 15.3 N
(b) M1 v² = u² + 2as with u=0, a=4.704, s=1.2
(b) A1 v² = 11.2896 (oe, ft on a from part (a))
(b) A1 v = 3.36 m/s cao
(b) Answer: v = 3.36 m/s
(c) M1 recognising new deceleration of A is μ g (friction only, no tension) = 2.94 m/s²
(c) A1ft correct value of deceleration, ft from μ used in part (a)
(c) M1 use v² = u² - 2as with u = 3.36 (ft from part (b)), v = 0
(c) A1ft s = 11.2896/5.88 (oe, ft on u from part (b))
(c) A1 s = 1.92 m cao
(c) Answer: 1.92 m
(d) B1 the string is modelled as light (massless) and inextensible
(d) B1 if the string had mass, its weight would affect the tension and the equations of motion in part (a), changing the acceleration and hence the speed used in part (c); the calculation in part (c) itself (friction only, string slack) would be unaffected once the string is slack, but the speed carried into it would differ
(d) Answer: String assumed light; a heavier string would change the acceleration and speed found earlier, altering the distance in part (c) even though the deceleration mechanism itself (friction only) stays the same.

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