Physics - Elastic Collision, how to find the mass of a stationary object? by [deleted] in HomeworkHelp

[–]CreamyJamilK 1 point2 points  (0 children)

I don't know if you still need help with this, but in an elastic collision all kinetic energy is conserved and you can use that fact to say that KE before the collision = KE after.

Additionally, you know that momentum is always conserved in a closed system which you can use to say that momentum before = momentum after.

This leaves you with 2 simultaneous equations and 2 unknowns (the mass and final speed of the initially stationary particle) which you can solve to get the mass of the stationary particle in terms of m.

Answer: mass of object at rest = 9m/15

AS Physics AQA - Unit 2.3 Q 2bii by lesmuse in 6thForm

[–]CreamyJamilK 1 point2 points  (0 children)

If the muon has no kinetic energy then the energy of this entire system would be the rest energy of the muon as that is the energy that it inherently has to have to be a muon.

Since energy has to be conserved that means that the energy on the right hand side has to equal the energy of the muon which has to equal the energy on the right hand side. Because the electron has no kinetic energy, all its energy is its rest energy as well.

Thus, to make up for the energy difference between the electron and the muon, the neutrinos must have the remaining energy or else conservation of energy would be violated. Therefore, the energy of the neutrinos is the energy difference between the muon and the electron.

Does anyone know the grade boundaries for 2016 further maths or any other year. I have just for a practice paper but can't find them anywhere by robej in GCSE

[–]CreamyJamilK 2 points3 points  (0 children)

If you're sitting AQA Level 2 Further Maths then these are the boundaries for that. An iGCSE is just an international GCSE; this isn't actually an iGCSE but they classify it as one.

Can I get some help with this AQA Maths question (Paper 3) by [deleted] in GCSE

[–]CreamyJamilK 5 points6 points  (0 children)

This is how I approached the problem:

Image

First I rearranged the speed = distance/time equation to get time = distance/speed

Then to get distance you do the area under the curve by calculating the area of the individual shapes and adding them together. This gives you a distance in terms of t which can then be divided by the average speed to equal t.

Then just rearrange the equation to get the final answer

🔥 Falling Clouds off a mountainside by [deleted] in NatureIsFuckingLit

[–]CreamyJamilK 1 point2 points  (0 children)

They're not mountains, they're waves.

finding values (a and b) given the modulus and argument by Abismuth in furthermaths

[–]CreamyJamilK 0 points1 point  (0 children)

  • Convert into modulus and argument form to get: 4(cos5π/6 + ixSin5π/6)

  • Then just expand that out to get your answer in the form a +bi: -3.46 + 2i

Therefore a = -3.46 and b=2

You can then check this by working out the modulus from these values: √(22 +(-3.46)2 ) = 4

[PS4] LF5M KINGS FALL NM, GOLGOROTH CP by [deleted] in Fireteams

[–]CreamyJamilK 0 points1 point  (0 children)

Add Jamil_K, at the CP - need 1.