**Question**

Two blocks A and B of masses 4 kg and 5 kg respectively are joined by a light inextensible string. The

blocks rest on a smooth plane inclined at an angle \(\alpha \)to the horizontal, where\( tan\alpha \frac{7}{24} \).The string is

parallel to a line of greatest slope of the plane with B above A. A force of magnitude 36 N acts on B,

parallel to a line of greatest slope of the plane (see diagram).**(i)** Find the acceleration of the blocks and the tension in the string.

**(ii)** At a particular instant, the speed of the blocks is 1 \(ms^{-1}\). Find the time, after this instant, that it

takes for the blocks to travel 0.65 m.

**Answer/Explanation**

**(i)** Apply Newton’s second law to either or to the system

Block A: \( T–4g\times \frac{7}{25}=4a\)

BlockB: \( 36-t-5g\times \frac{7}{25}=5a\)

system: \(36-5g\times \frac{7}{25}-4g\times 4g\times \frac{7}{25}=9a \)

Either solving the system for

a or solving a pair of simultaneous equations for either a or T a= \(1.2ms^{-2}\) t=16N

**(ii)\(** \left [ 0.65=1\times t+\frac{1}{2}\times 1.2t^{2} \right ]\) t = 0.5 s

**Alternative method for question (ii)**

\(\nu ^{2}=1^{2}+2\times 1.2\times 0.65 [v=1.6] and 0.65=\frac{1}{2}(1+v)t\) t = 0.5 s

### Question

A particle P of mass 8 kg is on a smooth plane inclined at an angle of 30Å to the horizontal. A forceof magnitude 100 N, making an angle of 1Å with a line of greatest slope and lying in the vertical plane containing the line of greatest slope, acts on P (see diagram).

(i) Given that P is in equilibrium, show that 1 = 66.4, correct to 1 decimal place, and find the normal reaction between the plane and P.

**Answer/Explanation**

(ii) Given instead that 1 = 30, find the acceleration of P.

100 cos θ = 8 g sin 30 → θ = 66.4

[R = 8 g cos 30 + 100 sin θ]

R = 161

3(ii) 100 cos 30 – 8g sin 30 = 8a

plane (3 terms) a = 5.83

### Question

Two particles A and B of masses 0.9 kg and 0.4 kg respectively are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two

inclined planes. The particles are initially at rest with A on a smooth plane inclined at angle \(\Theta ^{\circ}\) to the horizontal and B on a plane inclined at angle \(25 ^{\circ}\) to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes. A force of magnitude 2.5 N is applied to B acting down the plane (see diagram).

(i) For the case where 1 = 15 and the plane on which B rests is smooth, find the acceleration of B.

(ii) For a different value of \(\Theta\), the plane on which B rests is rough with coefficient of friction between the plane and B of 0.8. The system is in limiting equilibrium with B on the point of moving in the direction of the 2.5 N force. Find the value of \(\Theta\).

**Answer/Explanation**

(i) T – 0.9 g sin 15 = 0.9a

2.5 + 0.4 g sin 25 – T = 0.4a

1.3a = 1.86…

a = 1.43 m\( s^{–2}\)

(ii) F = 0.8 × 0.4g cos 25

2.5 + 0.4 g sin 25 – T – F = 0

T – 0.9 g sin θ = 0

θ = 8.2o\(^{\circ}\)

**Question**

Two light inextensible strings are attached to a particle of weight 25 N. The strings pass over two smooth fixed pulleys and have particles of weights A N and B N hanging vertically at their ends. The

sloping parts of the strings make angles of 30Å and 40Å respectively with the vertical (see diagram). The system is in equilibrium. Find the values of A and B.

**Answer/Explanation**

A cos 30 + B cos 40 = 25

A sin 30 = B sin 40

A = 17.1 B = 13.3

OR

\(\frac{25}{sin70}=\frac{A}{sin140}=\frac{B}{sin150}\)

A=17.1

B=13.3

**Question**

As shown in the diagram, a particle A of mass 0.8 kg lies on a plane inclined at an angle of 30Å to the horizontal and a particle B of mass 1.2 kg lies on a plane inclined at an angle of 60Å to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley P fixed at the top of the planes. The parts AP and BP of the string are parallel to lines of greatest slope of the respective planes. The particles are released from rest with both parts of the string taut.

(i) Given that both planes are smooth, find the acceleration of A and the tension in the string.

(ii) It is given instead that both planes are rough, with the same coefficient of friction, -, for both particles. Find the value of – for which the system is in limiting equilibrium.

**Answer/Explanation**

(i) [T – 0.8g sin 30 = 0.8a

1.2g sin 60 – T = 1.2a

1.2g sin 60 – 0.8g sin 30 = 2a]

For A T a − 4 0.8 =

For B \( \sqrt[6]{3}-T=10.4-T=1.2a\)

\(a=3\sqrt{3}-2=3.20ms^{-2}\)

T=\(\frac{12}{5}(1+\sqrt{3})=6.56N\)

(ii) RA = 0.8 cos30 = \(4\sqrt{3}\)

RB = 1.2 cos60 = 6

FA = 4 3 μ and FB = 6μ

12 sin 60 – 6μ – T = 0

orT – 8 sin 30 – 4√3 μ = 0

\(F_{A}=\sqrt[4]{3}\mu and F_{8}=6\mu\)

= 0.494

### Question

Two particles A and B of masses 2 kg and 3 kg respectively are connected by a light inextensible string. Particle B is on a smooth fixed plane which is at an angle of 18^{0} to horizontal ground. The string passes over a fixed smooth pulley at the top of the plane. Particle A hangs vertically below the pulley and is 0.45 m above the ground (see diagram). The system is released from rest with the string taut. When A reaches the ground, the string breaks.

Find the total distance travelled by B before coming to instantaneous rest. You may assume that B does not reach the pulley.

**Answer/Explanation**

Particle A: 2 g − T = 2a

Particle B: T – 3g sin 18 = T – 9.27 = 3a

System: 2g – 3g sin18 = 2g -9.27 = (2 + 3)a

a= 2.145898034

[5a = 10.72949017]

v^{2} = 2 × a × 0.45

v^{2} = 2 × 2.145898034 × 0.45 = 1.931308…

[v = 1.389715162]

T = 0, ±3 g sin 18 = 3g

[a = ± 3.0901699]

[0 = 1.93 – 2 × 3.09 × s] [s = 0.312]

Total distance moved by B = 0.45 + 0.312 = 0.762m

### Question

Two particles *A* and *B* have masses m kg and 0.1 kg respectively, where *m* > 0.1. The particles are

attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and

the particles hang vertically below it. Both particles are at a height of 0.9 m above horizontal ground

(see diagram). The system is released from rest, and while both particles are in motion the tension in

the string is 1.5 N. Particle B does not reach the pulley.

(a) Find m. [4]

(b) Find the speed at which A reaches the ground. [2]

**Answer/Explanation**

(a) 0.1 kg particle 0.1 0.1 Tga

m kg particle mg – T = ma

System mg – 0.1g = (m+0.1) a

Solve for m [a = 5]

m = 0.3

(b) v^{2} = 0 + 2 × 5 × 0.9

v= 3 m s^{–1}

**Question**

Two particles P and Q of masses 0.5 kg and m kg respectively are attached to the ends of a light

inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two

inclined planes. The particles are initially at rest with P on a smooth plane inclined at 30Å to the

horizontal and Q on a plane inclined at 45Å to the horizontal. The string is taut and the particles can

move on lines of greatest slope of the two planes. A force of magnitude 0.8 N is applied to P acting

down the plane, causing P to move down the plane (see diagram).

(a) It is given that m = 0.3, and that the plane on which Q rests is smooth.

Find the tension in the string.

(b) It is given instead that the plane on which Q rests is rough, and that after each particle has moved a distance of 1m, their speed is 0.6ms^{-1}\). The work done against friction in this part of the motion is 0.5 J.

**Answer/Explanation**

**Ans:**

- Attempt Newton’s 2nd law for either P, Q or the system.

For P: 0.8+0.5g sin 30 – T = 0.5a

For Q: T-0.3g sin 45=0.3a

System: 0.8+0.5g sin 30-0.3g sin 45 = 0.8a

Attempt to solve for T.

T=2.56 N (3sf) - KE and PE for m kg particle:

\(\frac{1}{2}m \times 0.36=0.18m\) and \(mg sin 45=5\sqrt{2}m\)

KE and PE for 0.5 kg particle:

\(\frac{1}{2} \times 0.5 \times 0.36 = 0.09 \) and 0.5g sin 30 = 2.5

Apply the work-energy equation to the system as:

PE loss + WD by 0.8N = KE gain +0.5

\(0.5g\times 1 \times sin 30-mg \times 1 \times sin 45 + 0.8 \times 1\)

m=0.374Alternative method for question 7(b)

KE and PE for m kg particle:

\(\frac{1}{2}m\times 0.36=0.18m\) and \(mg sin 45 = 5\sqrt{2}m\)

a = 0.18 and 3.3 – T =0.5(0.18) leading to T = 3.21

For m kg particle:

WD by T=KE gain + PE gain + 0.5

\(3.21\times1=\frac{1}{2}m \times 0.36 + mg sin 45+ 0.5\)

m=0.374Alternative method for question 7(b)

KE and PE for m kg particle:

\(\frac{1}{2}m \times 0.36=0.18m\) and mg sin 45 =\(5\sqrt{2}m\)

KE and PE for 0.5 kg particle

\(\frac{1}{2}\times 0.5 \times 0.36 = 0.09\) and 0.5g sin 30 = 2.5

Apply the work-energy equation to both particles as;

\(0.8\times 1+0.5gsin30=\frac{1}{2}\times0.5\times 0.36+T \times 1\)

and \(T\times 1=\frac{1}{2}m \times 0.36 + mg sin 45+0.5\)

\(0.8\times 1+0.5gsin30-\frac{1}{2}\times 0.5 \times 0.36=\frac{1}{2}m \times 0.36 + mg sin 45+0.5\)

m = 0.374

**Question**

Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible

string. Particle A is held on a smooth plane inclined at an angle of 1Å to the horizontal. The string passes over a small smooth pulley P fixed at the top of the plane, and B hangs freely 0.5 m above horizontal ground (see diagram). The particles are released from rest with both sections of the string

taut.**(i)** Given that the system is in equilibrium, find 1.

**(ii)** It is given instead that 1 = 20. In the subsequent motion particle A does not reach P and B remains at rest after reaching the ground.**(a)** Find the tension in the string and the acceleration of the system.

**Answer/Explanation**

**6(i)** ParticleA: T = 4 sinθ Particle B: T = 2 θ = 30

**6(ii)**(a)

A: T – 4 sin 20 = 0.4 a

B: 2 –T = 0.2 a System: 2 – 4 sin 20 = (0.4+ 0.2) T = 1.79 and a = 1.05

**6(ii)(b)**

v2 = 2× 1.053 × 0.5 = 1.053

v = 1.03 ms^{-2}

**6(ii)(c)**

Loss in KE = 1⁄2× 0.4× 1.053 = 0.2106

Gain in PE = 0.4× 10 ×d sin 20

1⁄2× 0.4× 1.053 = 0.4× 10×d sin 20

Total dist A moves up plane = 0.5+d = 0.654 m

**Question**

A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of 21Å to the

horizontal. The particle is kept in equilibrium by a force of magnitude P N acting parallel to a line of

greatest slope of the plane, as shown in the diagram. The coefficient of friction between the particle

and the plane is 0.3. Show that the least possible value of P is 0.470, correct to 3 significant figures, and find the greatest possible value of P.

**Answer/Explanation**

R = 0.6g cos 21 [= 5.60]

F = 0.3R = 1.8 cos 21 [= 1.68]

P + F = 6 sin 21[ = 2.15]

P = 2.15 – 1.68 = 0.470

P – F = 6 sin 21

P = 2.15 + 1.68 = 3.83

**Question**

Two particles P and Q, of masses 0.6 kg and 0.4 kg respectively, are connected by a light inextensible string. The string passes over a small smooth light pulley fixed at the edge of a smooth horizontal table. Initially P is held at rest on the table and Q hangs vertically (see diagram). P is then released. Find the tension in the string and the acceleration of Q

**Answer/Explanation**

[0.4g – T = 0.4a T = 0.6a

System equation 0.4g = (0.4 + 0.6)a]

\(a=4ms^{-2}\)

T = 2.4N

### Question

Blocks *P *and *Q*, of mass *m* kg and 5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a rough plane inclined at 35° to the horizontal. Block *P* is at rest on the plane and block *Q* hangs vertically below the pulley (see diagram). The coefficient of friction between block *P* and the plane is 0.2. Find the set of values of *m* for which the two blocks remain at rest. [6]

**Answer/Explanation**

Ans:

F = 0.2 × mg cos 35

5g – mg sin 35 – 0.2 mg cos 35

= 0

5g – Mg sin 35 + 0.2 Mg cos 35

= 0 A1

m = 6.78 or M = 12.2

\(6.78\leqslant mass\leqslant 12.2\)

### Question

Particles A and B, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string. Particle A is held at rest on a rough horizontal table with the string passing over a smooth pulley fixed at the edge of the table. The coefficient of friction between A and the table is 0.2. Particle B hangs vertically below the pulley at a height of 0.5 m above the floor (see diagram). The system is released from rest and 0.25 s later the string breaks. A does not reach the pulley in the subsequent motion. Find

** (i)** the speed of B immediately before it hits the floor, [9]

** (ii)** the total distance travelled by A. [3]

**Answer/Explanation**

Ans:

**Question**

A small block B of mass 0.25 kg is attached to the mid-point of a light inextensible string. Particles P and Q, of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of the string. The string passes over two smooth pulleys fixed at opposite sides of a rough table, with B resting in limiting equilibrium on the table between the pulleys and particles P and Q and block B are in the same vertical plane (see diagram).**(i)** Find the coefficient of friction between B and the table. Q is now removed so that P and B begin to move.

(ii) Find the acceleration of P and the tension in the part PB of the string.

**Answer/Explanation**

**(i) **Frictional force = μ × 0.25g

0.3g = 0.2g + μ0.25g

Coefficient of friction is 0.4

**(ii)** 0.2g – T = 0.2a or

T – 0.4 × 0.25g = 0.25a

T – 0.4 × 0.25g = 0.25a or

0.2g – T = 0.2a or

0.2g – μ0.25g = (0.2 + 0.25)a

Acceleration is \(2.22ms^{-1}\)Tension is 1.56N

**Question**

Small blocks A and B are held at rest on a smooth plane inclined at 30Å to the horizontal. Each is held in equilibrium by a force of magnitude 18 N. The force on A acts upwards parallel to a line of greatest slope of the plane, and the force on B acts horizontally in the vertical plane containing a line of greatest slope (see diagram). Find the weight of A and the weight of B.

**Answer/Explanation**

For A: right angle between 18 and R and

\(30^{\circ}\)opposite 18 or

WAsin\(30^{\circ} \)= 18 or

For B: right angle between 18 and W and \(30^{\circ}\)

opposite 18 or WB \(sin30^{\circ}\)

= 18cos \(30^{\circ}\)

For B: right angle between 18 and W and\( 30^{\circ}\)

opposite 18 or

WB \(sin30^{\circ}\)

= 18cos \(30^{\circ}\)or

For A: right angle between 18 and R and \(30^{\circ}\) opposite 18 or WAsin\(30^{\circ}\)= 18

Weight of A is 36N

and weight of B is 31.2N

### Question

** **

A light inextensible string has a particle A of mass 0.26 kg attached to one end and a particle B of mass 0.54 kg attached to the other end. The particle A is held at rest on a rough plane inclined at angle α to the horizontal, where sin \(\alpha =\frac{5}{13}\). The string is taut and parallel to a line of greatest slope of the plane. The string passes over a small smooth pulley at the top of the plane. Particle B hangs at rest vertically below the pulley (see diagram). The coefficient of friction between A and the plane is 0.2. Particle A is released and the particles start to move.

**(i)** Find the magnitude of the acceleration of the particles and the tension in the string. [6]

Particle A reaches the pulley 0.4 s after starting to move.

**(ii)** Find the distance moved by each of the particles. [2]

**Answer/Explanation**

Ans:

**Question**

Particles A and B, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. A is held at rest on a rough horizontal table with the string passing over a small smooth pulley at the edge of the table. B hangs vertically below the pulley (see diagram). The system is released and B starts to move downwards with acceleration 1.6 m\( s^{−2}\). Find

**(i)** the tension in the string after the system is released, **(ii)** the frictional force acting on A.

**Answer/Explanation**

**(i)** [0.2g – T = 0.2 × 1.6]

Tension is 1.68 N

**(ii) **T – F = 0.3 × 1.6

Frictional force is 1.2 N

**Question**

Particles P and Q, of masses 0.6 kg and 0.4 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a vertical

cross-section of a triangular prism. The base of the prism is fixed on horizontal ground and each of the sloping sides is smooth. Each sloping side makes an angle θ with the ground, where sin θ = 0.8.

Initially the particles are held at rest on the sloping sides, with the string taut (see diagram). The

particles are released and move along lines of greatest slope.**(i)** Find the tension in the string and the acceleration of the particles while both are moving.

The speed of P when it reaches the ground is 2 \(ms{−1}\). On reaching the ground P comes to rest and remains at rest. Q continues to move up the slope but does not reach the pulley.**(ii)** Find the time taken from the instant that the particles are released until Q reaches its greatest

height above the ground.

**Answer/Explanation**

** (i)** 0.6g × 0.8 – T = 0.6a and T – 0.4g × 0.8 = 0.4a

or (0.6 – 0.4)g × 0.8 = (0.6 + 0.4)a

Tension is 3.84 N or acceleration is \(1.6ms^{–2}\) Acceleration is 1.6 ms–2 or tension is 3.84 N

**(ii)** 2 = \(1.6t_{1}\) \((t_{1} \)= 1.25)

0 = \(2 – 0.8gt_{2}\) \((t_{2}\) = 0.25)

Time taken in 1.5 s