A2 Physics Unit 4June 2010 Multichoice

KS5 A2 PHYSICS 2450Mr D Powell

Raw Marks....

A =B =C = D = E = U =

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B

Ft = mv-mu F = (mv-mu)/t F = ∆mv/t

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C

Area under graph...ft 5kN x 40/2 = 100kNs-1

Then mv = momentum. mv/m = 50ms-1

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D

mv-mu v=-u This becomes... mv-(m-u) 2mv

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D

Only accelerating force is gravity in vertical plane.

Assuming air resistance is negligible.

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D

Using the idea that in a circle the velocity relates to pi and radius and period

Then sub out “v” from standard KE formula.

vT

rr

v

T

r

vT

f

2

2

2

2

2

22

2

22

2

2

2

2

4

2

2

1

2

1

T

mrKE

T

rmKE

T

rmKE

mvKE

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C

Simply apply formulae for velocity at place x.

339.07.09.09.8

12

2

22

22

v

xAfv

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A

and C is the potential energy!

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B

For A & D there is no change as T m or T k

By inspection leaves B or C TL=1.414T Tm=0.707T So period is double.

g

lT

k

mT

2

2

g

lT

g

lT

l

l

22

22

k

mT

k

mT

m

m

2

12

22

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?

Force is in direction as...

A

2r

GmMF

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A

W=mV W/m=V V = 8Jkg-1 / 10m V = 0.80 Jkg-1m-1 V =0.8 ms-2

V =0.8 Nkg-1

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?

...... B

vr

GM

r

mv

r

GmM

r

mv

r

GmMF

2

2

2

2

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D

20

2

20

21

4

d

QF

r

QQF

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A

Difference in charge is 20C spread over 120mm.

Points should cancel out 4/20th away from +4 C charge

So (120mm /20C) * 4C = 24mm

r

QV

04

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A r

QV

04

1

At point P distance by simple pythagoras.

Then difference between +Q and –Q for potential hill is 2Q.

+Q and +Q cancel out.

24

2

4

1

2

0

0

22

a

QV

r

QV

aaa

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B

Simply use formulae Sub in for “Q” to get It Then plug in numbers

10

2010

2

1

2

1

2

1

2

2

2

C

ItE

C

QE

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D

V must go down by 10% so plug into decay equ. V=IR as constant current so plug this in as well.

sCI

V

tRC

RC

t

V

V

eVV

o

RC

t

10549.0

1ln

9.0

1ln

ln

0

IRV

IRV

VV

/

9.00

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A

Simple decay curve on way down from exp equation. Reverse on way up

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C

Q=0.5Q after 10 seconds...

4.142ln

10

102ln

5.010

0

RC

RC

RC

QeQ

eQQ

RC

RC

t

o

o

QQ

eQ

Q

367.0

367.01

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A

F=BiL Hence direction of current creates opposite

effect If i -> 2i then effect doubled

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C

Using FBI. Electron is negative

From image electron moves in reverse to direction of current (second finger)

B-field is first finger

Hence C as not elliptical!

Cannot be D or A/B as curves

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B

Make sure you use pos->neg for current flow on second finger – rotate hand to make field N->S

Motion Downwards.

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B B – no as Ws-2=Js-3

C – certainly is by units

D – Fd = work or WD/D = F

A - ATm – sub into the equation and works. & F = BIL (of course!)

NATm

mCm

sN

s

CATm

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D

Faradays law Area gives flux change Grad of graph is

tt

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?

......

C

If speed increases then PD output goes up.

Also frequency of rotation increases according to formulae as shown.

BAN

t

tBAN

sin

sin

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C

Use Transformer Equation...

Np/Ns = Vp/Vs

Work out the unknown PD

Then work out input Power.

Then efficiency