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# Cambridge ENGAA question help watch

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1. Stuck on this question, not quite sure how you get to the answer at all!
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2. bump
3. Maybe use conservation of momentum to find the velocity of the alpha particle, then sub this into kinetic energy formula.

I will try it myself and see if theres any other help i can give
4. (Original post by Shaanv)
Maybe use conservation of momentum to find the velocity of the alpha particle, then sub this into kinetic energy formula.

I will try it myself and see if theres any other help i can give
I tried that and the answer I got was not even remotely close to any of the options
5. (Original post by Samii123)
I tried that and the answer I got was not even remotely close to any of the options
try again - I reduced the equation to just v^2 and E in the equation and went from there. you want to sub in values you know to get rid of speed of uranium as you don't need it
6. (Original post by bruh2132)
try again - I reduced the equation to just v^2 and E in the equation and went from there. you want to sub in values you know to get rid of speed of uranium as you don't need it
Well the correct answer is D but I'm struggling to figure out how the mass of Uranium ends up on the bottom
7. (Original post by Samii123)
Well the correct answer is D but I'm struggling to figure out how the mass of Uranium ends up on the bottom
you have simultaneous equations
Momentum:
MV=mv
and
0.5 M V^2 + 0.5 m v^2 = E

You know all the values of M you just have to eliminate V in the equation as you are only concerned about the energy of the Alpha so

V=4v/234

Put that into the kinetic energy equation and you eventually end up with 476(v^2)/234 = E
simplify and you'll get the answer
8. (Original post by bruh2132)
you have simultaneous equations
Momentum:
MV=mv
and
0.5 M V^2 + 0.5 m v^2 = E

You know all the values of M you just have to eliminate V in the equation as you are only concerned about the energy of the Alpha so

V=4v/234

Put that into the kinetic energy equation and you eventually end up with 476(v^2)/234 = E
simplify and you'll get the answer
I've got the answer now, thanks for the help!

how's everyone prepping? also is there an official thread for the ENGAA?
xx
10. (Original post by ndk123)

how's everyone prepping? also is there an official thread for the ENGAA?
xx
you still need help?
thanks

(Original post by erawein)
you still need help?
12. (Original post by ndk123)
thanks
3C) ok so, you differentiated to get the acceleration. remember maximum and minimum points?
so if you differentiate your function of acceleration and make it equal to zero and solve for t you should get 2T/3 or something and put it back in your acceleration equation to get the answer. note this gives you a minimum
3D) so you know that the gradient of the curve gives you acceleration. so which part of the curve is the steepest? that would be at the beginning where t = 0. put that in your acceleration equation to get the most positive acceleration.
3E) the area under the curve is the displacement? M1?
anyway integrate your velocity function. even though it says 2T the graph ends at T so put your limits as T and 0 to get the displacement
4D) firstly we know that centripetal acceleration goes towards the middle. so at the very highest of the loop do f=ma so N+mg = mv^2/r
Remember the normal reaction is down because the car is upside down and the weight obviously goes down as well. anyway the question previous to that had a equation for v sub that in for v in the f=ma equation.
now, for the car to stay on the loop N> 0
so N = mv^2/r - mg
in inequality terms mv^2/r -mg > 0
if you rearrange you should get the answer.

now this was me remembering what i had done, so if theres mistakes or you dont get it ask again, and ill thoroughly look into it. but hopefully ive explained well enough

13. I don't understand on this question, when you differentiate, why you don't differentiate (a^2)x as a product rule because when you do you don't get any of the answers but when you differentiate so that it becomes a^2 you can get the correct answer from it so I'm not sure why you differentiate it this way as I thought the product rule is meant to be applied here.
14. (Original post by Samii123)

I don't understand on this question, when you differentiate, why you don't differentiate (a^2)x as a product rule because when you do you don't get any of the answers but when you differentiate so that it becomes a^2 you can get the correct answer from it so I'm not sure why you differentiate it this way as I thought the product rule is meant to be applied here.
it’s a constant. it doesn’t change unlike x. it’s like saying x^3-2x^2 in that 2 is a constant.
15. (Original post by erawein)
it’s a constant. it doesn’t change unlike x. it’s like saying x^3-2x^2 in that 2 is a constant.
Wait no I understand it's only when functions of x are being multiplied by each other, I was getting confused because I was thinking about when I had to use the product rule during implicit differentiation with Xs and Ys but you would still include the dx/dy when differentiating the Y only then could you do that as a product rule.

16. One more question if anyone could help please
17. Bump
18. bump
19. Differentiate to find the turning points and figure out between which k values there'll be four intercepts.
20. (Original post by donnaseemchandra)
Differentiate to find the turning points and figure out between which k values there'll be four intercepts.
Yeah I did that and got x as -1 and 2 and then subbed them in to get -12 and 15 as k but the -12 is not part of the answer and I'm not sure why.

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