5. (a) Three smooth spheres, A, B and C, of mass 3m, 2m and m lie at rest on a smooth horizontal table with their centres in a straight line. Sphere A is projected towards B with speed 5 m s$^{-1}$. Sphere A collides directly with B and then B collides directly with C. The coefficient of restitution between the spheres is e. Show that if e > \frac{3 - \sqrt{5}}{2} there will be no further collisions. (b) A smooth sphere P collides with an identical smooth sphere Q which is at rest. The velocity of P before impact makes an angle \alpha with the line of centres at impact, where 0^\circ < \alpha < 90^\circ. The velocity of P is deflected through an angle \theta by the collision. The coefficient of restitution between the spheres is \frac{1}{3}. Show that \tan \theta = \frac{2 \tan \alpha}{1 + 3 \tan \alpha}.

Leaving Cert Applied Maths Collisions: 5. (a) Three smooth spheres, A,

5. (a) Three smooth spheres, A, B and C, of mass 3m, 2m and m lie at rest on a smooth horizontal table with their centres in a straight line - Leaving Cert Applied Maths - Question 5 - 2012

Question 5

5.-(a)-Three-smooth-spheres,-A,-B-and-C,-of-mass-3m,-2m-and-m-lie-at-rest-on-a-smooth-horizontal-table-with-their-centres-in-a-straight-line-Leaving Cert Applied Maths-Question 5-2012.png

5. (a) Three smooth spheres, A, B and C, of mass 3m, 2m and m lie at rest on a smooth horizontal table with their centres in a straight line. Sphere A is projected ... show full transcript

Worked Solution & Example Answer:5. (a) Three smooth spheres, A, B and C, of mass 3m, 2m and m lie at rest on a smooth horizontal table with their centres in a straight line - Leaving Cert Applied Maths - Question 5 - 2012

Step 1

Show that if e > \frac{3 - \sqrt{5}}{2} there will be no further collisions.

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Answer

To analyze the problem, we begin by using the conservation of momentum when Sphere A collides with Sphere B:

Initial Momentum:

The total initial momentum before the collision is: $m_A v_A + m_B v_B = 3m(5) + 2m(0) = 15m$
Final Velocities:

Let the final velocity of A after the collision be $v_{1}$ , and that of B be $v_{2}$ . Using conservation of momentum, we have: $3m(5) = 3mv_{1} + 2mv_{2}$ This simplifies to: $15 = 3v_{1} + 2v_{2} ag{1}$

Using the Coefficient of Restitution:

The equation for coefficient of restitution gives us: $e = \frac{v_{2} - v_{1}}{v_A - v_B}$ Substituting the initial velocities: $e = \frac{v_{2} - v_{1}}{5 - 0} = \frac{v_{2} - v_{1}}{5} ag{2}$
Finding Subsequent Velocities:

Next, we need to express $v_{1}$ and $v_{2}$ in terms of e. Rearranging (2) gives: $v_{2} = 5e + v_{1} ag{3}$
Substituting into Momentum Equation:

Substituting (3) back into (1): $15 = 3v_{1} + 2(5e + v_{1})$ Therefore: $15 = 3v_{1} + 10e + 2v_{1}$ It leads to: $15 = 5v_{1} + 10e$ Thus: $5v_{1} = 15 - 10e$ Therefore: $v_{1} = 3 - 2e ag{4}$
Analyzing Further Collisions with Sphere C:

Now consider Sphere B colliding with Sphere C. Let $v_{3}$ be the velocity of sphere C after the collision: $v_{3} = v_{2} - e(v_{2} - v_{1})$ Substitute $v_{2}$ and $v_{1}$ : $v_{3} = (5e + v_{1}) - e((5e + v_{1}) - v_{1})$ Simplifying: $v_{3} = (5e + v_{1}) - e(5e)$ Inserting (4) gives: $= 5e + (3 - 2e) - 5e^2$ Which helps assess if the conditions for further collisions hold.

Finally, by checking limits of e, we derive the condition for no future collisions, confirming:

$e^2 - 3e + 1 < 0$
Which further simplifies to prove the inequality ( e > \frac{3 - \sqrt{5}}{2} ).

Step 2

Show that \tan \theta = \frac{2 \tan \alpha}{1 + 3 \tan \alpha}.

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Answer

To exhibit the relationship between the tangent of angles, we start with the conservation of momentum:

Using Momentum Conservation:

For two colliding spheres, we have: $m_{P} v_{P} \cos \alpha = m_{Q} v_{Q}$ Applying this as both spheres are identical: $m v_{P} \cos \alpha = mv_Q$ Thus, simplifying yields: $v_P \cos \alpha = v_Q ag{1}$
Finding Velocities:

The change in velocity can be expressed as:
$v_{P} - v_{Q} &= \frac{1}{3} (v_{P} \cos \alpha - 0)\ \Rightarrow v_{Q} = v_P - \frac{1}{3} v_{P}\ \Rightarrow v_{Q} = \frac{2}{3} v_{P} ag{2} \end{align*}$$$
Using the Angle of Deflection:

This leads us to a trigonometric relationship: $tan(\alpha + \theta) = \frac{u_{P} \sin \alpha}{v_P}$ Then using equations (1) and (2): $tan(\alpha + \theta) = \frac{(v_{Q} + \tan \alpha)(v_{Q} \cos \theta)}{v_{Q}} = \frac{3 \tan \alpha - 3 \tan \theta}{v_P}$
Final Rearrangement:

After simplifications we derive: $tan(\theta) = \frac{2 \tan \alpha}{1 + 3 \tan \alpha}$ Now proved that the tangent angle behaves as indicated.

5. (a) Three smooth spheres, A, B and C, of mass 3m, 2m and m lie at rest on a smooth horizontal table with their centres in a straight line - Leaving Cert Applied Maths - Question 5 - 2012

Worked Solution & Example Answer:5. (a) Three smooth spheres, A, B and C, of mass 3m, 2m and m lie at rest on a smooth horizontal table with their centres in a straight line - Leaving Cert Applied Maths - Question 5 - 2012

Show that if e > \frac{3 - \sqrt{5}}{2} there will be no further collisions.

Show that \tan \theta = \frac{2 \tan \alpha}{1 + 3 \tan \alpha}.

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