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
Question 17
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.
The combined mass of the bugg... show full transcript
Worked Solution & Example Answer: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
Step 1
Find R.
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Answer
To find the resistance force R acting on the roller-skater, we start by applying Newton's second law to the buggy and roller-skater system.
Assume the total force acting on the buggy is given by the driving force minus the total resistance. Hence, we write:
Fnet=Fdriving−Fresistance
Where:
Combined mass of the buggy and driver = 410 kg.
Driving force = 300 N.
Total resistance on buggy = 140 N.
Acceleration = 0.2 m/s².
Using Newton's second law:
Fnet=mimesa
Thus, we have:
Fnet=410imes0.2=82N
Setting the forces equal gives us:
82N=300N−(140N+R)
Rearranging the equation:
R=300N−140N−82N
Therefore:
R=78N
Step 2
Find the tension in the rope.
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Answer
To find the tension T in the rope, we again apply Newton's second law on the roller-skater.
The net force acting on the roller-skater is:
Fnet=T−R
Where:
Mass of the roller-skater = 72 kg.
Acceleration = 0.2 m/s².
Thus, we can write:
Fnet=72imes0.2=14.4N
Setting the forces gives:
14.4N=T−78N
Rearranging the equation yields:
T=14.4N+78N
Therefore:
T=92.4N
Step 3
State a necessary assumption that you have made.
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Answer
A necessary assumption made in this problem is that the rope connecting the buggy and the roller-skater is massless, and that friction or air resistance does not significantly affect the system.
Step 4
Determine whether the roller-skater will stop before reaching the stationary buggy.
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Answer
To determine whether the roller-skater stops before reaching the stationary buggy after releasing the rope at speed of 6 m/s, we will first calculate the deceleration experienced due to the resistance force.
We can write the equation of motion as:
T−R=ma
Where:
R = 78 N.
T = 0 (no tension after releasing).
Mass of the roller-skater = 72 kg.
Hence, we apply Newton's law to find acceleration:
0−78N=72imesa
Solving for a gives:
a = -rac{78}{72} = -1.083 ms^{-2}
To find the stopping distance (s), we can use the kinematic equation:
v2=u2+2as
Where:
Final velocity v = 0 (when the roller-skater stops).
Initial velocity u = 6 m/s.
Using the calculated a:
0=(6)2+2(−1.083)s
Rearranging gives:
s = rac{36}{2 imes 1.083} \\ s = 16.63 m
Since the stopping distance (16.63 m) is less than the distance to the buggy (20 m), the roller-skater will indeed stop before reaching the stationary buggy.
Step 5
Explain the change in motion that the driver noticed.
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Answer
Upon releasing the rope, the roller-skater continues to experience resistance while moving forward. As the roller-skater loses the tension that was aiding their motion, a combination of their initial momentum and the resisting forces results in a slowing motion. The driver notices that the buggy begins to change speed, as it continues to travel with an initially applied driving force, leading to the buggy eventually coming to a stop at a distance of 20 m from the roller-skater's release point.