1. Q 3 Revisited Why did the two different methods produce different results?

2. The question was inconsistent! Initial momentum was 0.117 kgms -1 One of the momentum vectors was at an angle of 50 degrees. The other vector was at 90 degrees to that one. Now the triangle is fixed! All quantities can be calculated. There is no freedom to make up results.

3. Calculate what the velocities should have been. Initial momentum was 0.117 kgms -1 50 degrees. 90 degrees The masses have a mass of 0.065 kg. 40 degrees

4. Trigonometry. Initial momentum was 0.117 kgms -1 50 degrees. 90 degrees Sin 40 = p 2 / 0.117 p 2 = 0.117 x sin 40 = 0.0752 kgms -1 40 degrees p2p2 p1p1 Cos 40 = p 1 / 0.117 p 1 = 0.117 x cos 40 = 0.0896 kgms -1 Pythagoras : sqrt of (0.0752 2 + 0.0896 2 ) = 0.11697 – Agrees!!

5. Velocities p 2 = 0.0752 kgms -1 = 0.065kg x V 2 V 2 = p 2 / 0.065 = 1.156 ms -1 = 1.2 ms -1 (2 s.f) p 1 = 0.0896 kgms -1 = 0.065kg x V 1 V 1 = p 1 / 0.065 = 1.378 ms -1 = 1.4 ms -1 (2 s.f)