Dynamics and Space Mock Assessment 2.1
1 A snowboarder sets off from rest and accelerates uniformly at 4.1 m s-2 down a straight run.
After 3.9 seconds the skier reaches a speed of 16 m s-1 and maintains this speed for a further 11 s.
(a) State what is meant by an acceleration of 4.1 m s-2
The velocity increases by 4.1 m s-1 each second.
(b) Calculate the distance travelled by the skier during the first 3.9 s.
s  ut  1 / 2at 2
s  0  0.5 x 4.1x3.9 x3.9
s  31m
(c) Explain, in terms of the forces acting on the skier, why the skier
continues to travel at 16 m s-1 after the first 3.9 s.
The forces are balanced
1
2
A golf ball is hit by a club and leaves the club with a momentum of 3·90 kgms-1. The
club is in contact with the ball for a time of 0·210 s.
(a) Calculate the average force exerted by the club on the ball.
Ft  mv  mu
Fx0.210  3.90
3.90
0.210
F  18.6 N
F
(b) (b)The graph shows how the force, F, exerted on the ball by the club varies with
time t.
F
0
i.
t
(c) State what quantity is represented by the area under a force-time graph.
Impulse
(d) The experiment is now repeated using a soft squash ball. This ball receives the
same momentum as the golf ball.
i. Copy the graph and draw an additional line to show how the force varies with time for the
sponge ball.
hard
Lower F max, longer time
soft
ii. You must label each line clearly.
2
3
A football is kicked and leaves the ground with an initial velocity of 13 m s-1 at an angle
of 35o to the horizontal as shown.
13m s-1
35o
(a) State what happens to the magnitude of the vertical velocity of the football as it climbs to its
maximum height.
Vertical velocity (magnitude) decreases to 0 m s-1.
NOT
velocity decreases
decreases on its own
Assume that air resistance is negligible.
(b) Calculate the horizontal component of the initial velocity of the ball.
Vh=13cos35
=11 ms-1
(c) Calculate the vertical component of the initial velocity of the ball.
Vv=13sin35
=7 ms-1
(d)
Calculate the maximum height reached by the ball.
v 2  u 2  2as
0  7.456 x7.456  2 x(9.8) xs
0  55.5919  19.6 xs
 55.5919  19.6 xs
 55.5919
s
 19.6
s  2.836m
The total time of flight is 1·62 s.
(e) Calculate the horizontal distance travelled by the ball in this time.
d  vxt
d  10.648 x1.62
17.25m
(f)
The ball is kicked again with the same initial speed but at a smaller angle to the horizontal. Predict
how this affects the maximum height reached by the football.
If angle to horizontal is less
=>vertical component Vv=13sinX will be less
=> will reach a smaller height
3
3
(a)
Explain what is meant by time dilation.
Time dilation is a difference in a time interval as measured by a stationary observer and a
moving observer (or equivalent).
(a) A spacecraft leaves Earth and travels at a constant speed of 0·4c to its destination. An astronaut on
board the spacecraft measures the time to reach its destination to be exactly 2·5 years. Calculate the
time taken for the journey as measured by an observer on Earth.
t 
t 
t
v
1  
c
2 .5
2
1  0.4 
2
2 .5
1  0.16
2 .5
t 
0.84
2 .5
t 
0.91
t   2.7 years
t 
4
5
An ambulance is approaching a stationary pedestrian.
(a) The siren on the ambulance emits sound with a frequency of
840 Hz. The speed of sound in air is 3∙40 × 102 m s-1.
The ambulance is travelling at a constant speed of 22 m s-1. Calculate the frequency
heard by the pedestrian as the ambulance approaches
 v 

f 0  f s 
v

v
s


 340 
f 0  840

 340  22 
 900 Hz
.
(b) Explain in terms of wavefronts why the sound heard by the pedestrian is different
from the frequency emitted by the siren.
Frequency higher because wave length shorter (ahead of the ambulance)
Diagram showing wavefronts closer together ahead of the ambulance than behind it.
Wavefronts bunched
together in front
Wavefronts spread out
behind
(c) Light from a distant galaxy exhibits redshift.
What is meant by the term redshift?
Light from objects moving away is shifted to longer wavelengths.
5
6
State one piece of evidence that supports the Big Bang theory.
Any correct piece of evidence with no wrong physics eg
 red shift
 Cosmic Microwave Background radiation
.
The peak wavelength, λp, of the emitted radiation is inversely proportional to the kelvin
temperature, T, of the hot object as shown by the following graph.
(a) The peak wavelength of a distant star is 714 nm so 1/λp = 1·4 × 106 (m-1)
Use the graph to determine the temperature of the star
Temperature, T = 3600 K (from graph)
(b) Show by calculation which of the following options
describes the relationship
between the temperature, T, in K, and peak wavelength, λp, in m
gradient 
(6000  0)
 3 103 (K m)
6
(2 10  0)
(a) T = 3 × 10-3 λp
(b) T = 3 × 10-3 ×
1
λp
1
(c) T = 333 × λp
(d) T = 333 λp
6