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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??

energy is increased to 4E. Why is the new de Broglie wavelength λ/2??

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#2

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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??

**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??

KE = 0.5mv^2 => 4KE => 2v

λ = h/mv

As v has doubled, λ will half.

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(Original post by

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.

**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.

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#4

(Original post by

Thanks for the quick reply, I'm not too good at Physics.

**HK4**)Thanks for the quick reply, I'm not too good at Physics.

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(Original post by

Don't worry, and feel free to PM me any questions.

**britishtf2**)Don't worry, and feel free to PM me any questions.

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 ?

Answer is 𝑚𝑣^2/𝑟𝑔

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

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#6

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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 ?

Answer is 𝑚𝑣^2/𝑟𝑔

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

**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 ?

Answer is 𝑚𝑣^2/𝑟𝑔

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

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.

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(Original post by

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.

**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.

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#8

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Thanks a lot, http://filestore.aqa.org.uk/resource...-74081-SQP.PDF its question 29, I thought I'd attached an image sorry.

**HK4**)Thanks a lot, http://filestore.aqa.org.uk/resource...-74081-SQP.PDF its question 29, I thought I'd attached an image sorry.

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.

Answer:

Mg = (mv^2)/r

M = (mv^2)/rg

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#9

(Original post by

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.

**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.

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#10

(Original post by

How did you learn first question, book doesnt tell you and my physics teacher is trash

**flyingthin**)How did you learn first question, book doesnt tell you and my physics teacher is trash

GPE = mgh = KE = 0.5mv^2

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

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