A smooth sphere A, of mass 3m, moving with speed u, collides directly with a smooth sphere B, of mass 5m, which is at rest - Leaving Cert Applied Maths - Question 5 - 2013

The impulse $I$ imparted can be expressed as follows:

1. From the definition of impulse, we have:

   $I = m(v_1 - u) = 5m(v_2 - 0)$

2. Setting the magnitudes equal gives:

   $2mu = 5m(v_2) \Rightarrow v_2 = \frac{2u}{5}$

3. Substitute $v_2$ back into the equation we have:

   $v_2 = \frac{3u(1+e)}{8} = \frac{2u}{5}$

4. Cross-multiplying results in:

   $3u(1+e)\cdot5 = 16u\Rightarrow 15(1+e) = 16 \Rightarrow 15 + 15e = 16 \Rightarrow 15e = 1 \Rightarrow e = \frac{1}{15}$

Using the formula for the coefficient of restitution:

1. We know:

$v^2 = u^2 + 2as$

With $s = h$:

$v^2 = 0 + 2gh \Rightarrow v = \sqrt{2gh}$

2. Considering the rebound, the new height is \frac{h}{4}:

$0 = e\sqrt{2gh} - 2g\frac{h}{4}$

3. Thus:

$0 = e\sqrt{2gh} - \frac{gh}{2} \Rightarrow e = \frac{\frac{gh}{2}}{\sqrt{2gh}} = \frac{1}{2}$

Using the same formula:

1. If thickness of paper is given as 2.5 cm, we get:

$0 = e\sqrt{2gh} - 2g\frac{h}{9}$

2. Thus:

$e\sqrt{2gh} = \frac{2gh}{9} \Rightarrow e = \frac{1}{3}$

Continuing with:

1. When rebound is $\frac{h}{16}$:

$0 = e\sqrt{2gh} - 2g\frac{h}{16}$

2. From this, we have:

$e(\sqrt{2gh}) = \frac{gh}{8} \Rightarrow e = \frac{1}{4}$

Given:

$e = k(\text{thickness}),\: k = \frac{1}{2} \Rightarrow \frac{1}{4} = \frac{1}{2}(\text{thickness}) \Rightarrow thickness = \frac{1}{2\cdot \frac{1}{4}} = \frac{10}{3} cm = 3.33 cm$