# A level Physics simple MCQ

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#1
An electron has a kinetic energy E and a de Broglie wavelength λ. The kinetic
energy is increased to 4E. Why is the new de Broglie wavelength λ/2??
0
3 years ago
#2
(Original post by HK4)
An electron has a kinetic energy E and a de Broglie wavelength λ. The kinetic
energy is increased to 4E. Why is the new de Broglie wavelength λ/2??
Kinetic energy increased implies increased speed (as mass stays constant, assuming no relativistic effects).
KE = 0.5mv^2 => 4KE => 2v

λ = h/mv
As v has doubled, λ will half.
1
#3
(Original post by britishtf2)
Kinetic energy increased implies increased speed (as mass stays constant, assuming no relativistic effects).
KE = 0.5mv^2 => 4KE => 2v

λ = h/mv
As v has doubled, λ will half.
Thanks for the quick reply, I'm not too good at Physics.
0
3 years ago
#4
(Original post by HK4)
Thanks for the quick reply, I'm not too good at Physics.
Don't worry, and feel free to PM me any questions.
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#5
(Original post by britishtf2)
Don't worry, and feel free to PM me any questions.
An object falls freely from rest. After falling a distance d its velocity is v. What is its velocity after it has fallen a distance 2d? Answer is √2 v

An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5 m s −1 . What is the efficiency of the system? Answer is 50%

A string passes through a smooth thin tube. Masses m and M are attached to the ends of the string. The tube is moved so that the mass m travels in a horizontal circle of constant radius r and at constant speed v. What is equal to M ?

Here are the other question are didn't quite get in the paper, if you or anyone else could have a go at explaining that would be great
0
3 years ago
#6
(Original post by HK4)
An object falls freely from rest. After falling a distance d its velocity is v. What is its velocity after it has fallen a distance 2d? Answer is √2 v

An electric motor of input power 100 W raises a mass of 10 kg vertically at a steady speed of 0.5 m s −1 . What is the efficiency of the system? Answer is 50%

A string passes through a smooth thin tube. Masses m and M are attached to the ends of the string. The tube is moved so that the mass m travels in a horizontal circle of constant radius r and at constant speed v. What is equal to M ?

Here are the other question are didn't quite get in the paper, if you or anyone else could have a go at explaining that would be great
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)

2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)

3) See posts 7 and 8.

1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)

2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf

3, ans) See post 8.
0
#7
(Original post by britishtf2)
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)

2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)

3) Have you got a diagram for this? i.e. Is the tube vertical/horizontal?

1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)

2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf

3, ans) Will do once I know what the set up is.
Thanks a lot, http://filestore.aqa.org.uk/resource...-74081-SQP.PDF its question 29, I thought I'd attached an image sorry.
0
3 years ago
#8
(Original post by HK4)
Thanks a lot, http://filestore.aqa.org.uk/resource...-74081-SQP.PDF its question 29, I thought I'd attached an image sorry.
No problem.
The weight of mass M is providing the centripetal force of the mass m. It is the only force providing it, so it is equal to it.
Try to work from that.

Mg = (mv^2)/r
M = (mv^2)/rg
0
8 months ago
#9
(Original post by britishtf2)
1) Equate energies, and let d = h
change in mgh = change in 0.5mv^2 (ignoring pos/neg)
-Try it from here. I will post the full solution at the bottom for you (goes for all of them)

2) Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
(See below for full)

3) See posts 7 and 8.

1, ans)
change in mgh = change in 0.5mv^2
v^2 = 2gh
v = root(2gh)
Change in height doubles and 2g stays the same, so v increases by a factor of root(2)

2, ans)
Efficiency = 100%*(Useful energy out)/(Total energy in)
Consider 1 second:
Energy in = 100J, as Power = 100J/s
Useful energy out = Ke+Change in mgh per second
Ke = 0.5*10*0.5^2 = 1.25
Change in mgh in one second = mg*delta(h) = mg*vt = 10*9.8*0.5*1 = 49
49+1.25 = 50.25
Efficiency = 100%* 50.25/100 = 50.25% = 50% 2sf

3, ans) See post 8.
How did you learn first question, book doesnt tell you and my physics teacher is trash
0
4 months ago
#10
(Original post by flyingthin)
How did you learn first question, book doesnt tell you and my physics teacher is trash
Hello. Sorry for the (extremely) slow reply; I haven't logged into TSR in months and I hope it's not too late/this is still of some help to you. You can "equate energies" because energy is conserved (assuming no destruction/creation of mass), so you just have to figure out the forms of energies that are involved. When you are dropping something from a certain height, the gravitational potential energy (GPE) is converted into kinetic energy (KE) (ignoring conversion into heat energy via air resistance). So,

GPE = mgh = KE = 0.5mv^2

Hope this was of help, or you managed to figure it out yourself.
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