AQA A-Level Physics Year 2
Student Book Answers. Chapter 1
ASSIGNMENT 1
A1 The total lateral friction on the tyres to just avoid the car going off at a tangent is equal to the required
centripetal force:
F =
m v
2
r
=
690´(120´10
3 ¸3600)
2
70
=10952 =11000 N
A2 The asymmetric cur...
Chapter 1
ASSIGNMENT 1
A1 The total lateral friction on the tyres to just avoid the car going off at a tangent is equal to the required
centripetal force:
2
mv 690 ´(120 ´103 ¸3600) 2
F = = =10952 =11000 N
r 70
A2 The asymmetric curved shape and angle of a racing car aerofoil makes moving air flow faster below it than
above, so the air pressure is lower below the aerofoil. This creates a downward force on the aerofoil pushing
the car downwards.
v 2
A3 At the Abbey bend, g-force = centripetal acceleration = r
acceleration due to gravity g
(120 ´103 ¸3600)2 ¸70
= =1.6
9.81
So the g-force typically experienced by a driver at the Abbey bend is 1.6g.
ASSIGNMENT 2
A1 a. Estimate from the diagram of the beam length from the centre to the back of the chair 5 m.
2
æ 2p ö
2 4p 2
Centripetal acceleration a =r w =r ´ç ÷ =5 ´ 2 m s- 2
èT ø 100
b. The normal contact force exerted by the seat on the rider’s back provides the centripetal force = mass
of rider ´ centripetal acceleration. Estimate of rider’s mass 60 kg.
Normal contact force 60 ´2 =120 N
A2 a. There is nothing to provide a centripetal force on them, so they tend to move tangentially until they hit
the side of the drum.
2
2
æ2 3000 ö
b. Centripetal acceleration a =r w =r ´(2pf ) =0.3 ´ç 2p ´ ÷ =30000 m s –2
è 60 ø
c. Using the estimate of the mass of a swimming costume 50 g, the normal contact force, N, on the
inside surface on the swimming costume is given by
N =m a 0.05 ´30000 =1500 N
ASSIGNMENT 3
A1
2
, 2
m v
A2 a. m g +N =
r
60 ´8.82
b. N = - 60 ´9.81=341 N
5
2
mv
A3 N – m g =
r
60 ´152
N = +(60 ´9.81) =2090 N
9
2
m v
A4 At the top of the loop m g + N =
r
2
m v
If N = 0, m g = , which gives
r
v = r g = 5 ´9.81 =7.0 m s- 1
PRACTICE QUESTIONS
1a. i.
2
mv
1a. ii. m g +T =
r
Hence
0.05 ´3 2
T = - (0.05 ´9.81) =0.26 N
0.6
1b. i.
2
mv
1b. ii. T -mg =
r
1c. The string is most likely to snap when the mass is at the bottom of its path since the tension in the string is at
its greatest at this point.
2a. Although the aircraft has a constant speed, its direction is changing. Since velocity is a vector quantity, if the
direction changes the velocity must also be changing. If there is a velocity change there must be some
acceleration since acceleration is equal to rate of change of velocity.
2b. When the aircraft is banked, it rotates about its centre of mass so that the lift force on the wings is no longer
acting in the vertical direction. The horizontal component of the lift creates a resultant force towards the
centre of the circle which provides the centripetal force. (See figure 16 in the Student Book.)
3
, v 2 (280 ´103 ¸3600)2
2c. i. The banking angle can be found from tan = = =0.2056 , which gives the
rg 3000 ´9.81
banking angle, = 12°.
2
v (280 ´103 ¸3600) 2
2c. ii. centripetal acceleration a = = =2.0 m s- 2
r 3000
mv 2
3a. The centripetal force on the electron is given by F = =8.2 ´10 - 8 N. Rearranging gives the electron’s
r
r 0.053 ´10 - 9
speed v = ´8.2 ´10 - 8 = ´8.2 ´10 - 8 =2.2 ´10 6 m s - 1
m 9.11´10 - 31
v 2.2 ´10 6
3b. The angular speed w = = = 4.151´1016 = 4.2 ´1016 rad s - 1
r 0.053 ´10 - 9
2p 2p
3c. The time for one orbit T = = =1.5 ´10 - 16 s
w 4.151´1016
2 D 2
4. C (Centripetal force: F =mr w = m
2
2pf = 2mDp2f 2 )
2pr 2 ´p ´1.5 ´1011
5. D (Speed v = = =3.0 ´104 m s- 1)
T 365 ´24 ´3600
æ1 2ö
6. A (The speed of the object is uniform so the kinetic energy ç m v ÷ must not change)
è2 ø
2p 15 ´2 ´p p
7. A (The speed of the man, v =r w =r ´ = = m s- 1 )
T 50 ´60 100
m g
8. C (When the mass is about to slip, the centripetal force m r w 2 is equal to the maximum frictional force .
2
m g g
Hence m r w 2 = . Rearranging gives = )
2 2r
4
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