Edexcel M1 2005 June — Question 7 13 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2005
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeConnected particles via tow-bar on horizontal surface
DifficultyModerate -0.3 This is a standard M1 mechanics question involving Newton's second law applied to connected particles and subsequent motion with deceleration. Parts (a)-(c) require straightforward application of F=ma and kinematic equations with clear problem setup. Part (d) requires understanding of vertical force components but is still routine. Slightly easier than average A-level due to the structured, multi-part nature guiding students through the solution with standard techniques.
Spec3.02d Constant acceleration: SUVAT formulae3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution

\includegraphics{figure_4} Figure 4 shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is at an angle of \(15°\) to the road. The lorry and the car experience constant resistances to motion of magnitude 600 N and 300 N respectively. The lorry's engine produces a constant horizontal force on the lorry of magnitude 1500 N. Find
  1. the acceleration of the lorry and the car, [3]
  2. the tension in the towbar. [4]
When the speed of the vehicles is \(6 \text{ m s}^{-1}\), the towbar breaks. Assuming that the resistance to the motion of the car remains of constant magnitude 300 N,
  1. find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest. [4]
  2. State whether, when the towbar breaks, the normal reaction of the road on the car is increased, decreased or remains constant. Give a reason for your answer. [2]

Question 7:
AnswerMarks
7600
300 T T 1500
(a) Lorry + Car: 2500a = 1500 – 300 – 600
a = 0.24 m s–2
(b) Car: T cos 15 – 300 = 900a OR Lorry: 1500 – T cos 15 – 600 = 1600a
Sub and solve: T ≈ 534 N
(c) 300 Deceleration of car = 300/900 = 1/3 m s–1
Hence 62 = 2 x 1/3 x s ⇒ s = 54 m
(d) Vertical component of T now removed
AnswerMarks
Hence normal reaction is increasedM1 A1
A1
(3)
M1 A1
↓ ↓
M1 A1
(4)
M1 A1
M1 A1
(4)
M1
A1 cso
(2)
Question 7:
7 | 600
300 T T 1500
(a) Lorry + Car: 2500a = 1500 – 300 – 600
a = 0.24 m s–2
(b) Car: T cos 15 – 300 = 900a OR Lorry: 1500 – T cos 15 – 600 = 1600a
Sub and solve: T ≈ 534 N
(c) 300 Deceleration of car = 300/900 = 1/3 m s–1
Hence 62 = 2 x 1/3 x s ⇒ s = 54 m
(d) Vertical component of T now removed
Hence normal reaction is increased | M1 A1
A1
(3)
M1 A1
↓ ↓
M1 A1
(4)
M1 A1
M1 A1
(4)
M1
A1 cso
(2)
\includegraphics{figure_4}

Figure 4 shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is at an angle of $15°$ to the road. The lorry and the car experience constant resistances to motion of magnitude 600 N and 300 N respectively. The lorry's engine produces a constant horizontal force on the lorry of magnitude 1500 N. Find

\begin{enumerate}[label=(\alph*)]
\item the acceleration of the lorry and the car, [3]
\item the tension in the towbar. [4]
\end{enumerate}

When the speed of the vehicles is $6 \text{ m s}^{-1}$, the towbar breaks. Assuming that the resistance to the motion of the car remains of constant magnitude 300 N,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{2}
\item find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest. [4]
\item State whether, when the towbar breaks, the normal reaction of the road on the car is increased, decreased or remains constant. Give a reason for your answer. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2005 Q7 [13]}}