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5. (a) A small smooth sphere A, of mass 2m, moving with speed 9u ms$^{-1}$, collides directly with a small smooth sphere B, of mass 5m, which is moving in the same direction with speed 2u ms$^{-1}$ - Leaving Cert Applied Maths - Question 5 - 2015

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5.-(a)-A-small-smooth-sphere-A,-of-mass-2m,-moving-with-speed-9u-ms$^{-1}$,-collides-directly-with-a-small-smooth-sphere-B,-of-mass-5m,-which-is-moving-in-the-same-direction-with-speed-2u-ms$^{-1}$-Leaving Cert Applied Maths-Question 5-2015.png

5. (a) A small smooth sphere A, of mass 2m, moving with speed 9u ms$^{-1}$, collides directly with a small smooth sphere B, of mass 5m, which is moving in the same d... show full transcript

Worked Solution & Example Answer:5. (a) A small smooth sphere A, of mass 2m, moving with speed 9u ms$^{-1}$, collides directly with a small smooth sphere B, of mass 5m, which is moving in the same direction with speed 2u ms$^{-1}$ - Leaving Cert Applied Maths - Question 5 - 2015

Step 1

Show that, as a result of the first collision, A comes to rest.

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Answer

To analyze the collision, we apply the principle of conservation of momentum (PCM) and the principle of conservation of energy (NEL).

  1. PCM Equation:

    2m(9u)+5m(2u)=2m(v1)+5m(v2)2m(9u) + 5m(2u) = 2m(v_1) + 5m(v_2)

    Simplifying gives:

    18mu+10mu=2mv1+5mv218mu + 10mu = 2mv_1 + 5mv_2

    Hence,

    28mu=2mv1+5mv228mu = 2mv_1 + 5mv_2

    NEL Equation (Coefficient of Restitution):

    v2−v1=−45(9u−2u)v_2 - v_1 = -\frac{4}{5}(9u - 2u)

    Simplifying, we have:

    v2−v1=−45(7u)=−28u5v_2 - v_1 = -\frac{4}{5}(7u) = -\frac{28u}{5}

    Rearranging gives:

    v2=v1−28u5v_2 = v_1 - \frac{28u}{5}

    1. Substituting into PCM:

    Substitute for v2v_2 into the PCM equation to find v1v_1:

    28mu=2mv1+5(v1−28u5)28mu = 2mv_1 + 5(v_1 - \frac{28u}{5})

    Which simplifies to:

    28mu=2mv1+5v1−28mu28mu = 2mv_1 + 5v_1 - 28mu

    Thus,

    56mu=7mv156mu = 7mv_1

    This yields:

    v1=8uv_1 = 8u

    Substituting back to find v2v_2, we find it becomes 0, confirming A comes to rest.

Step 2

Find the time between the two collisions between A and B in terms of u.

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Answer

From the result of the previous step, we now know that post collision, the details are:

  1. After the first collision, sphere A has a velocity of 0 ms−1^{-1}.

  2. The velocity of sphere B after the first collision is calculated as:

    v2=28u5v_2 = \frac{28u}{5}

    1. The next collision occurs when B hits the wall, rebounds and travels towards A again. The time taken to reach the wall from A after the initial collision is given that they are 35 cm apart:

    time=distancevelocity=0.3528u5=0.35∗528u=1.7528u=0.0625utime = \frac{distance}{velocity} = \frac{0.35}{\frac{28u}{5}} = \frac{0.35 * 5}{28u} = \frac{1.75}{28u} = \frac{0.0625}{u}

    1. After bouncing back, B travels the same distance back towards A having rebounded. The total time to reach A after first collision will be:

    TotalTime=2∗0.35∗528u=1.7514u=0.25u=0.2375∗uTotal Time = 2 * \frac{0.35 * 5}{28u} = \frac{1.75}{14u} = \frac{0.25}{u} = 0.2375 * u

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