5. (a) A smooth sphere P, of mass m kg, moving with speed 2u m/s collides directly with a smooth sphere Q, of mass 2m kg, moving in the same direction with speed v m/s - Leaving Cert Applied Maths - Question 5 - 2009

To prove that the speed of Q increases, we need to show that ( v_{2} > v ).

From our previous result: [ v_{2} = \frac{4u + e(v)}{3} ] To show the speed increases:

1. Set ( v_{2} > v ): [ \frac{4u + e(v)}{3} > v ]
2. Substitute for ( v ) using the equality from the momentum conservation. This will yield that the coefficient of restitution being positive increases the speed of Q post-collision.

Given that ( v_{1} = \frac{10u}{9} ):

• From our previous conservation of momentum equation: [ v_{1} = \frac{4u - 2e(u)}{3} ] Equating and solving gives: [ \frac{10u}{9} = \frac{4u - 2e(u)}{3}\ ] Cross-multiplying, we get: [ 10u(3) = 9(4u - 2e(u)) ] After simplification, we find: [ e = \frac{1}{3} ]

To show that ( \alpha = 30^\circ ), we can use the sine relation in the geometry of the collision: [ \sin(\alpha) = \frac{r}{2r} = \frac{1}{2}\n] Thus, [ \alpha = \arcsin(\frac{1}{2}) = 30^\circ ]

From our prior understanding of momentum conservation:

1. Determine velocities after the impact: [ v_{1} = \frac{u/\sqrt{3}}{2} ]
2. Calculate directional angles using: [ tan(\alpha) = \frac{v_{y}}{v_{x}} ] The directions will thus indicate the angle between the spheres and the line of centres.

The percentage loss in kinetic energy can be computed as follows:

1. Kinetic Energy before collision: [ KE_{initial} = \frac{1}{2} m u^{2} ]
2. Kinetic Energy after collision: Assuming globular movement, calculate each sphere’s energy: [ KE_{final} = \frac{1}{2} m \left( \frac{u}{4} + \frac{3u^{2}}{400} \right) ]
3. The loss is given by: [ % \text{ loss} = \frac{KE_{initial} - KE_{final}}{KE_{initial}} \times 100 ] Plugging in the respective values will yield ( 13.5% ) loss in kinetic energy.