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 Pure - Question 17 - 2018 - Paper 2

Using the same principles, we can find the tension T in the rope. The forces acting on the roller-skater can be summarized as:

$T - R = m imes a$

Where:

• T is the tension we want to calculate.
• R is the resistance force we found earlier (63.6 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:

$T - 63.6 = 72 imes 0.2$

Calculating the right side gives:

$T - 63.6 = 14.4$

Thus,

$T = 14.4 + 63.6 = 78 N$

Therefore, the tension in the rope is 78 N.

One assumption made is that the rope connecting the buggy and the roller-skater is massless or does not have any significant mass that affects the forces involved.

Another assumption made is that there is no friction between the roller-skater's wheels and the ground.

To determine whether the roller-skater stops before reaching the buggy, we first find her deceleration after releasing the rope. The net force acting on her can be calculated as:

$F_{net} = -R$

Substituting for R: $F_{net} = -63.6 N$

Using Newton's second law: $F_{net} = m imes a$

The mass is 72 kg, hence:

$-63.6 = 72 imes a$

Solving for a gives: $a = -0.883 m/s^2$

The roller-skater's initial speed, right before she releases the rope, is 6 m/s. To find out how far she will travel before stopping, we can use the equation of motion:

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

Where:

• v = final velocity (0 m/s, when she stops),
• u = initial velocity (6 m/s),
• a = -0.883 m/s²,
• s = distance travelled.

Thus,

$0 = (6)^2 + 2(-0.883)s$

Solving for s:

$s = \frac{(6)^2}{2(0.883)} = 20.4 m$

Since she stops after 20.4 m and the buggy is 20 m away, the roller-skater will not stop before reaching the stationary buggy.

The driver noticed a change in motion of the buggy because when the roller-skater released the rope, the resistance experienced by the roller-skater decreased. As a result, the tension in the rope was lost, which abruptly altered the forces acting on both the buggy and roller-skater. The buggy continued to experience a net forward force from the driving force, while the roller-skater began to decelerate due to her resistance. This difference in acceleration led to the buggy slowing down and coming to a stop.