M1 June 2013 (R) Q7
7.
A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is inclined at an angle \(\theta\) to the road, as shown in Figure 4. The vehicles are travelling at 20 m s\(^{-1}\) as they enter a zone where the speed limit is 14 m s\(^{-1}\). The truck’s brakes are applied to give a constant braking force on the truck. The distance travelled between the instant when the brakes are applied and the instant when the speed of each vehicle is 14 m s\(^{-1}\) is 100 m.
The constant braking force on the truck has magnitude \(R\) newtons. The truck and the car also experience constant resistances to motion of 500 N and 300 N respectively. Given that \(\cos\theta = 0.9\), find
| Scheme | Marks |
|---|---|
| Use of \(v^2 = u^2 + 2as\) | M1 |
| \(14^2 = 20^2 - 2a \times 100\) | A1 |
| Deceleration is 1.02(m s\(^{-2}\)) | A1 |
| (3) |
Notes
M1 for a complete method to produce an equation in \(a\) only.
First A1 for a correct equation.
Second A1 for 1.02 (ms\(^{-2}\)) oe. must be POSITIVE.
| Scheme | Marks |
|---|---|
| Horizontal forces on the car: \(\pm T\cos\theta - 300 = 750 \times -1.02 = -765\) \(T = -1550/3\) | M1A2 f.t. |
| The force in the tow-bar is 1550/3, 520 (N) or better (allow –ve answer) | A1 |
| (4) |
Notes
M1 for considering the car ONLY horizontally to produce an equation in \(T\) only, with usual rules. i.e. correct no. of terms AND \(T\) resolved:
\(\pm T\cos\theta - 300 = 750 \times -1.02\)
A2 ft on their \(a\) for a correct equation (300 and \(a\) must have same sign); -1 each error (treat cos 0.9 as an A error)
A1 for 1550/3 oe, 520 or better (N) N.B. Allow a negative answer.
| Scheme | Marks |
|---|---|
| Horizontal forces on the truck: \(\pm T\cos\theta - 500 - R = 1750 \times -1.02\) | M1A2 f.t. |
| Braking force \(R = 1750\) (N) | A1 |
| (4) | |
| (11 marks) |
Notes
M1 for considering the truck ONLY horizontally to produce an equation, with usual rules. i.e. correct no. of terms AND \(T\) resolved:
\(\pm T\cos\theta - 500 - R = 1750 \times -1.02\)
A2 ft on their \(T\) and \(a\) for a correct equation (500, \(a\) and \(R\) must have same sign); -1 each error (treat cos 0.9 as an A error)
A1 for 1750 (N).
OR
M1 for considering the whole system to produce an equation in \(R\) only, with usual rules. i.e. correct no. of terms.
A2 ft on their \(a\) for a correct equation (\(a\) and \(R\) must have same sign) -1 each error
A1 for 1750 (N).
N.B. If 300 and 500 are given separately, penalise any sign errors only ONCE.
ALT
| Whole system: \(800 + R = 2500 \times 1.02\) | M1A2 f.t. |
| \(R = 1750\) | A1 |