June 2023 Paper 2 Q19

AQACurrent spec12 marksConnected Particles Resolving Forces

19 A wooden toy comprises a train engine and a trailer connected to each other by a light, inextensible rod.

The train engine has a mass of 1.5 kilograms.
The trailer has a mass 0.7 kilograms.

A string inclined at an angle of 40° above the horizontal is attached to the front of the train engine.

The tension in the string is 2 newtons.

As a result the toy moves forward, from rest, in a straight line along a horizontal surface with acceleration 0.06 m s⁻² as shown in the diagram below.

Toy train engine pulling a trailer along a horizontal surface, with acceleration 0.06 m s^−2 to the right and a string force of 2 N at 40° above the horizontal attached to the front of the engine

As it moves the train engine experiences a total resistance force of 0.8 N

(a) Show that the total resistance force experienced by the trailer is approximately 0.6 N [4 marks]

(b) At the instant that the toy reaches a speed of 0.5 m s⁻¹ the string breaks.

As a result of this the train engine and trailer decelerate at a constant rate until they come to rest, having travelled a distance of \(h\) metres.

It can be assumed that the resistance forces remain unchanged.

(i) Find the tension in the rod after the string has broken. [4 marks]

(ii) Find \(h\) [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Resolves the 2 N force to obtain either \(2\cos 40\) AWRT 1.53 or \(2\sin 40\) AWRT 1.29 May be seen on the diagram.	M1	1.1a
Uses Newton’s 2nd Law to form a four term equation for the whole system. This may be seen with total resistance equivalent to \(0.8 + R\) or Uses Newton’s 2nd Law to form a three term equation for the trailer or a four term equation for the engine. Condone one incorrect sign.	M1	3.3
Obtains a fully correct equation for the whole system. or Obtains two fully correct equations for the train engine and the trailer. For example: \(2\cos 40 - T - 0.8 = 0.09\) \(T - R = 0.042\)	A1	3.3
Completes a reasoned argument to show that the given value for \(R\) is approximately 0.6 AG	R1	2.1
	(4)

Typical solution

Use \(F = ma\) for system

\[2\cos 40 - (0.8 + R) = 2.2(0.06)\]\[1.53 - 0.8 - R = 0.132\]\[R \approx 0.6\text{ N}\]

Mark scheme (b)

Scheme	Marks	AO
(i) Forms an equation of motion without a driving force for the engine or the combined system. PI by \(a = \dfrac{7}{11}\) or \(-\dfrac{7}{11}\)	M1	3.3
Obtains one of the following: \(\pm(T - 0.6) = 0.7a\) or \(\mp(0.8 + T) = 1.5a\) or \(\pm(0.8 + 0.6) = 2.2a\) Allow use of \(a = -\dfrac{7}{11}\)	M1	3.4
Obtains a correct pair of equations of motion for the trailer and the engine which are both moving in the same direction. or Obtains \(a = -\dfrac{7}{11}\) and one fully correct equation of motion involving \(T\)	A1	1.1b
Finds \(T\) AWFW [0.15 , 0.16]	A1	1.1b
	(4)
(ii) Obtains \(a = -\dfrac{7}{11}\) AWRT \(-0.64\) Condone missing or incorrect units	B1	3.1b
Selects an appropriate equation of constant acceleration to find \(s\) and substitutes \(u\) = 0.5 , \(v\) = 0 and their \(a\) Do not accept \(a = -g\)	M1	1.1a
Obtains required distance. AWRT 0.2 ISW	A1	1.1b
	(3)

Typical solution

(i)

\[-0.8 - T = 1.5a\]\[T - 0.6 = 0.7a\]\[T = \frac{17}{110}\text{ N}\]

(ii)

\[a = -\frac{7}{11}\]\[0 = 0.5^2 + 2ah\]\[h = \frac{11}{56} \approx 0.20\]

Scheme	Marks	AO
States one appropriate modelling assumption about the rod Accept Rod is rigid OE	E1	3.5b
	(1)
	(12 marks)