A buggy is pulling a roller-skater, in a straight line along a horizontal road, by means of a connecting rope as shown in the diagram - AQA - A-Level Maths Mechanics - Question 17 - 2018 - Paper 2

To find the tension T in the rope, we again apply Newton's second law to the roller-skater alone:

The net force acting on the roller-skater can be expressed as:

$T - R = ma$

Where:

• T is the tension in the rope.
• R is the total resistance force calculated previously (R = 78 N).
• m is the mass of the roller-skater (72 kg).
• a is the acceleration (0.2 m s^-2).

Substituting the known values into the equation:

$T - 78 = 72 imes 0.2$

Thus,

$T - 78 = 14.4$

Rearranging gives:

$T = 14.4 + 78 = 92.4 N$

Thus, the tension in the rope is:

$T = 92.4 N$

A necessary assumption made in this scenario is that the rope has no mass and is inextensible, meaning it does not stretch. This ensures that the tension is the same throughout the length of the rope and that the forces can be treated as acting simultaneously.

To determine whether the roller-skater will stop before reaching the buggy, we need to calculate the stopping distance after the rope is released. The skater continues with an initial speed of 6 m/s and experiences a resistance force of 78 N. Using the equation of motion:

$v^2 = u^2 + 2as$

Where:

• v is the final velocity (0 m/s as the skater comes to a stop),
• u is the initial velocity (6 m/s),
• a is the acceleration (negative due to resistance, calculated as)

a = - rac{R}{m}

Substituting:

a = -rac{78}{72} = -1.083 m/s^2

Now substituting into the motion equation:

$0 = (6)^2 + 2(-1.083)(s)$

Solving for s gives:

$36 = 2(-1.083)s$ s = rac{36}{2.166} ext{ approximately } 16.6 m

Since the braking distance of approximately 16.6 m is less than the 20 m distance to the buggy, it can be concluded that the roller-skater will stop before reaching the stationary buggy.

The change in motion that the driver noticed is attributed to the absence of tension in the rope once the roller-skater released it. Prior to the release, the tension in the rope directly affected the motion of the buggy by contributing to the acceleration. When the skater released the rope, the only forces acting on the buggy were the driving force and the total resistance. As a result, this sudden change in force led to a noticeable deceleration of the buggy, causing it to come to a stop at a distance from point A.