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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

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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.

  1. Assume the total force acting on the buggy is given by the driving force minus the total resistance. Hence, we write:

    Fnet=FdrivingFresistanceF_{net} = F_{driving} - F_{resistance}

    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=mimesaF_{net} = m imes a

    Thus, we have:

    Fnet=410imes0.2=82NF_{net} = 410 imes 0.2 = 82 N

    Setting the forces equal gives us:

    82N=300N(140N+R)82 N = 300 N - (140 N + R)

    Rearranging the equation:

    R=300N140N82NR = 300 N - 140 N - 82 N

    Therefore:

    R=78NR = 78 N

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.

  1. The net force acting on the roller-skater is:

    Fnet=TRF_{net} = T - R

    Where:

    • Mass of the roller-skater = 72 kg.
    • Acceleration = 0.2 m/s².

    Thus, we can write:

    Fnet=72imes0.2=14.4NF_{net} = 72 imes 0.2 = 14.4 N

    Setting the forces gives:

    14.4N=T78N14.4 N = T - 78 N

    Rearranging the equation yields:

    T=14.4N+78NT = 14.4 N + 78 N

    Therefore:

    T=92.4NT = 92.4 N

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.

  1. We can write the equation of motion as:

    TR=maT - 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:

    078N=72imesa0 - 78 N = 72 imes a

    Solving for a gives:

    a = - rac{78}{72} = -1.083 ms^{-2}

  2. To find the stopping distance (s), we can use the kinematic equation:

    v2=u2+2asv^2 = u^2 + 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)s0 = (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.

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