A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope - Edexcel - A-Level Maths Mechanics - Question 6 - 2006 - Paper 1

To find the tension in the tow-rope, we can consider the forces acting on the trailer, which is also in motion. The equation for the trailer is:
[ T - F_{resistance} = m_{trailer} \cdot a ]
Where:

• Resistance force on the trailer, ( F_{resistance} = 280 \text{ N} )
• Mass of the trailer, ( m_{trailer} = 700 \text{ kg} )
• Acceleration (using the result from part (a)), ( a = 1.25 \text{ m/s}^2 )

Substituting the known values into the equation:
[ T - 280 = 700 \cdot 1.25 ]
[ T - 280 = 875 \Rightarrow T = 875 + 280 = 1155 \text{ N} ]

After the tow-rope breaks, the only forces acting on the car will be the driving force and resistance. Since the problem states that the driving force and resistances remain unchanged, we first calculate the new acceleration:

Using the net force as calculated earlier:
[ F_{net} = 2380 - 630 = 1750 \text{ N} ]

Now substituting it into the formula to find the new acceleration:
[ a_{new} = \frac{F_{net}}{m_{car}} = \frac{1750}{1400} = 1.25 \text{ m/s}^2 ]

Now we can use kinematic equations to find the distance covered in the first 4 seconds after the break. Using the formula:
[ s = ut + \frac{1}{2} a t^2 ]
Where:

• Initial velocity ( u = 12 ext{ m/s} )
• Time ( t = 4 ext{ s} )
• Acceleration ( a = 1.25 ext{ m/s}^2 )

Plugging the values in:
[ s = 12 \cdot 4 + \frac{1}{2} \cdot 1.25 \cdot 4^2 ]
[ s = 48 + \frac{1}{2} \cdot 1.25 \cdot 16 ]
[ s = 48 + 10 = 58 \text{ m} ]

The assumption that the tow-rope is inextensible implies that the accelerations of both the car and trailer are the same while the rope is intact. As such, when determining the forces acting on the car and trailer, they can be modeled as a single system. This means the tension in the rope does not change instantaneously under load, allowing us to apply consistent equations of motion throughout the scenarios considered.