5. (a) A sphere, of mass m and speed u, impinges directly on a stationary sphere of mass 3m - Leaving Cert Applied Maths - Question 5 - 2010

First, we calculate the initial kinetic energy (K.E.):

$K.E._{initial} = \frac{1}{2}mu^2$

After the collision with $e = \frac{1}{4}$, we find:

• $v_1$: $v_1 = \frac{u(1 - \frac{1}{4})}{4} = \frac{3u}{16}$
• $v_2$: $v_2 = \frac{u(1 + 3 \cdot \frac{1}{4})}{4} = \frac{5u}{16}$

Now compute the final kinetic energies:

$K.E._{final} = \frac{1}{2}m \left( \frac{3u}{16} \right)^2 + \frac{1}{2}(3m) \left( \frac{5u}{16} \right)^2$

This yields: $K.E._{final} = \frac{m}{2} \left( \frac{9u^2}{256} \right) + \frac{3m}{2} \left( \frac{25u^2}{256} \right) = \frac{m}{2} \left( \frac{84u^2}{256} \right) = \frac{21mu^2}{128}$

The loss in kinetic energy is:

$Loss = K.E._{initial} - K.E._{final} = \frac{mu^2}{2} - \frac{21mu^2}{128} = \frac{64mu^2}{128} - \frac{21mu^2}{128} = \frac{43mu^2}{128}$

The percentage loss is:

$Percentage Loss = \frac{Loss}{K.E._{initial}} \times 100 = \frac{\frac{43mu^2}{128}}{\frac{mu^2}{2}} \times 100 = \frac{43}{64} \times 100 = 67.1875\%$

Using conservation of momentum and the component velocities:

• For horizontal: $m(6) + km(2) = mv_1 + km(v_2)$
• For vertical: $0 + 0 = mv_2 + km(v_3)$

From the equations, we have:

1. $6 + 2k = v_1 + kv_2$
2. $0 = v_2 + 4k$

From here, we can express $v_2$ in terms of $k$: $v_2 = -4k$

At this point: $6 + 2k = v_1 - 4k$ This yields: $v_1 = 6 + 6k$

Using the coefficient of restitution, we set up the equation:

e = k = \frac{v_2 - v_1}{u} Substituting gives: $e = \frac{-4k - (6 + 6k)}{6} = \frac{-10k - 6}{6} = -\frac{5}{3}$ rounded to the nearest whole number is usefule in expressing e in terms of k.

From the previously derived equation, we know: [ e = \frac{3k + 1}{2 + 4k} ]
In order to satisfy the inequality for e (which must be less than or equal to 1), we can solve for k:

to ensure $\frac{3k + 1}{2 + 4k} \\le 1$:

1. $3k + 1 \leq 2 + 4k$ This simplifies to:
2. $k \geq -1$ and since $k$ is a constant proposed that,