A bead, A, of mass 0.1 kg is threaded on a long straight rigid wire which is inclined at sin^{−1} \(\left ( \frac{7}{25} \right )\) to the horizontal. A is released from rest and moves down the wire. The coefficient of friction between A and the wire is μ. When A has travelled 0.45m down the wire, its speed is 0.6 m s^{−1}.

Another bead, B, of mass 0.5 kg is also threaded on the wire. At the point where A has travelled 0.45 m down the wire, it hits B which is instantaneously at rest on the wire. A is brought to instantaneous rest in the collision. The coefficient of friction between B and the wire is 0.275.

### (a) Question

Show that – = 0.25.

**Answer/Explanation**

0.6^{2} = 0 + 2 a × 0.45

a = 0.4

\(R = 0.1g\times cos\alpha = 0.1g\times \frac{24}{25} = 0.1g\times cos 16.3^{\circ}\left [ R=\frac{24}{25}=0.96 \right ]\)

\(0.1g\times \times \frac{7}{25} – F = 0.1g\times 0.4\left [ 0.28 – F = 0.04 \rightarrow F = 0.24 \right ]\)

\(F = \mu \times 0.1g\times \frac{24}{25}\left [ F = \frac{24\mu }{25} = 0.96\mu \right ]\)

µ = 0.25

### (b) Question

Find the time from when the collision occurs until A collides with B again.

**Answer/Explanation**

Ans:

0.1 × 0.6 = 0.5v

v = 0.12

For B 0.5g × \(\frac{7}{25}- 0.275 \times 0.5g \frac{24}{25} = 0.5a \left [ leading to a = 0.16 \right ]\)

\(s_{A} = 0 + \frac{1}{2}\times 0.4t^{2}\) \(s_{B} = 0.12t + \frac{1}{2}\times 0.16t^{2} \)

For both sA and s_{B} and attempt to solve s_{A} = s_{B} to find t

Required time is t = 1 s

A particle of mass 12 kg is stationary on a rough plane inclined at an angle of 25^{0} to the horizontal. A force of magnitude P N acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35.

### (a) Question

Draw a sketch showing the forces acting on the particle.

**Answer/Explanation**

Correct 4 force diagram

### (b) Question

Find the least possible value of P.

**Answer/Explanation**

Ans:

Attempt to resolve forces parallel to the plane

P + F = 12g sin 25 [=50.7]

R =12g cos 25 [= 108.8]

F = 120 cos 25 × 0.35 [= 38.1]

P + 38.1 = 120 sin 25

P = 12.6

**Question**

A block of mass 5kg is being pulled along a rough horizontal floor by a force of magnitude X N acting at \(30^o\) above the horizontal (see diagram). The block starts from rest and travels 2m in the first 5s of its motion.

(a) Find the acceleration of the block

(b) Given that the coefficient of friction between the block and the floor is 0.4, find X.

The block is now placed on a part of the floor where the coefficient of friction between the block and the floor has a different value. The value of X is changed to 25, and the block is now in limiting equilibrium.

(c) Find the value of the coefficient of friction between the block and this part of the floor.

**Answer/Explanation**

**Ans:**

- \([2=\frac{1}{2}\times a \times 25]\)

\(a=0.16 ms^{-2}\) - R=5g – X sin 30

X cos30 -F=5a

X cos 30-0.4(5g-X sin 30)=5 \times 0.16\)

X=19.5 (3sf) - R=(5g-25 sin 30)[R = 37.5]

F = 25 cos 30 \([F=\frac{25\sqrt{3}}{2}]\)

\(\mu=\frac{F}{R}=0.577\) (3sf)

**Question**

A particle Q of mass 0.2 kg is held in equilibrium by two light inextensible strings PQ and QR. P is a fixed point on a vertical wall and R is a fixed point on a horizontal floor. The angles which strings PQ and QR make with the horizontal are \(60^o\) and \(30^o\) respectively (see diagram)

Find the tensions in the two strings.

**Answer/Explanation**

**Ans:**

For attempting to resolve forces in either direction.

\(T_P cos 60= T_R cos 30\)

\(T_p sin 60=T_R sin 30+0.2g\)

Attempt to solve simultaneously for either tension.

\(T_P=3.46N\) and \(T_R=2 N\)

**Alternative method for question 3**

\(\frac{T_P}{sin 60}=\frac{T_R}{sin150}=\frac{0.2g}{sin150}\)

One pair correct

Equation all correct

Solve for \(T_P\) or \(T_R\)

\(T_P=3.46N\) and \(T_R=2 N\)

### Question

The diagram shows a ring of mass 0.1 kg threaded on a fixed horizontal rod. The rod is rough and the

coefficient of friction between the ring and the rod is 0.8. A force of magnitude T N acts on the ring

in a direction at 30Å to the rod, downwards in the vertical plane containing the rod. Initially the ring

is at rest.

(a) Find the greatest value of T for which the ring remains at rest. [4]

(b) Find the acceleration of the ring when T = 3. [3]

**Answer/Explanation**

Ans

(a) Resolving forces in either direction

R = T sin30 + 0.1g, F = T cos30

T cos30 = 0.8 (T sin30 + 0.1g)

T = 1.72 (1.7166…)

(b) R = 3sin30 + 0.1g

3 cos30 – 0.8(3sin30 + 0.1g) = 0.1a

a = 5.98 ms^{–2} (5.9807…)

**Question**

Two particles A and B, of masses m kg and 0.3 kg respectively, are attached to the ends of a light

inextensible string. The string passes over a fixed smooth pulley and the particles hang freely below

it. The system is released from rest, with both particles 0.8 m above horizontal ground. Particle A reaches the ground with a speed of\( 0.6 ms^{-1}\)

**(i)** Find the tension in the string during the motion before A reaches the ground.

**(ii)** Find the value of m.

**Answer/Explanation**

**(i) **0.6^{2}=0+2a× 0.8\

a= 0.225

T-0.3g=0.3a

T=3.07N(3.0675N)

(ii) mg–T= ma, m(10–0.225)= 3.0675

m= 0.314 kg (0.31381…)

### Question

A car of mass 1800 kg is towing a trailer of mass 400 kg along a straight horizontal road. The car and

trailer are connected by a light rigid tow-bar. The car is accelerating at 1.5 m s^{−2}. There are constant

resistance forces of 250 N on the car and 100 N on the trailer.** (a) Find the tension in the tow-bar. **

(b) Find the power of the engine of the car at the instant when the speed is 20 m s^{−1}.

**Answer/Explanation**

(a) [T – 100 = 400 × 1.5]

T = 700 N

(b)F – 250 – 100 = 2200 × 1.5 (F = 3650 N)

(**M1** for using Newton’s second law for the system or for the car using the result from **2(a))**

For use of power =*Fv*

73 000 W or 73 kW

### Question

Two particles A and B, of masses 0.8 kg and 0.2 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. The particles hang vertically. The system is released from rest. Show that the acceleration of A has magnitude\( 6 m s^{−2}\) and find the tension in the string.

**Answer/Explanation**

[T – 2 = 0.2a 8 – T = 0.8a]

System is 0.8g – 0.2g = (0.2 + 0.8)a

and T = 2(0.2)(0.8)g/(0.8 + 0.2)

Attempt to solve for a or T

a = 6 T = 3.2

**Question**

A particle P of mass 1.6 kg is suspended in equilibrium by two light inextensible strings attached to points A and B. The strings make angles of 20Å and 40Å respectively with the horizontal (see diagram).

Find the tensions in the two strings.

**Answer/Explanation**

\(T_{A} sin 20\) + T_{B} sin 40 = 16

T_{A} cos 20 =\(T_{B}cos40\)

T_{A}=14.2N

T_{B}=17.4N

\(\frac{16}{Sin120}=\frac{T_{B}}{Sin130}\)

\(\frac{16}{Sin120}=\frac{T_{A}}{Sin110}\)

\(T_{A}=14.2N\)

\(T_{B}=17.4N\)