Transcript Impulse

Key Areas covered
• Force-time graphs during contact of
colliding objects.
• Impulse found from the area under a
force-time graph.
• Equivalence of change in momentum
and impulse.
• Newton’s third law of motion.
What we will do today
• State what is meant by impulse.
• Investigate force-time graphs and use
these to calculate impulse.
Impulse
An object is accelerated by a force, F, for a
time, t.
The unbalanced force is given by:
Fun = ma
= m(v- u)
t
= mv – mu
t
Unbalanced force
=
= change in momentum
time
rate of change of
momentum
Impulse = change in momentum
Impulse = force x time
Impulse = Ft = mv – mu
Units of Impulse are kgms-1 or Ns.
Impulse is a vector quantity, so DIRECTION
is important.
The concept of impulse is useful in situations
where the force is not constant and acts for
a very short period of time.
An example of this is when a golf ball is hit
by a club.
During contact, the
unbalanced force
between the club
and the ball varies
with time (often ms)
as shown in the graph
opposite.
F
0
t
Making use of impulse
• We can make use of impulse to help with safety
features.
• Re-arranging the equation to F = mv-mu / t shows
that to decrease the force, you have to increase the
time of contact.
• This is why cars have a crumple zone and why helmets
have padding inside.
• These safety features increase the time of contact
which decreases the average force applied
(decreasing the damage on the human body).
• This can be expressed in graph form.
• Note the change in momentum does not
change (mass, initial velocity and final velocity
are all the same).
2003 Qu: 5
Chevrolet Bel Air vs. 2009
Chevrolet Malibu crash test. YouTube
Newton’s third law
• For every action there is an equal and
opposite reaction.
• This holds true for impulse, the force
created during contact between the two
object (i.e. golf club and ball) is equal.
Bouncing Balls
This effect can also be seen when using
balls of different materials.
A hard ball, such as a basket ball,
deforms only a little.
Click on video to start. Click on green area to move to next slide.
A softer ball, such as a squash ball,
deforms more and so the time of
contact is longer.
Click on video to start. Click on green area to move to next slide.
• Tiger Woods Slo Mo + Close up of
Ball.mpg
In practical situations the force is not
constant, but comes to a peak and then
decreases.
Impulse = Area under a Force-time graph
In any collision involving impulse, the
unbalanced force calculated is always the
average force and the maximum force
experienced would be greater than the
calculated average value.
Note: time is often given in ms, this must
be converted to seconds (x10-3)
Example 1
A snooker ball of
mass 0.2 kg is
accelerated from
rest to a velocity
of 2 ms-1 by a
force from the
cue which lasts
for 50 ms.
What size of
force is exerted
by the cue?
Solution
u = 0, v = 2 ms-1, m = 0.2 kg,
t = 50 ms = 0.05 s, F = ?
Ft
F x 0.05
F
F
=
=
=
=
mv – mu
(0.2 x 2) – 0
0.4 / 0.05
8N
Example 2
A tennis ball of mass
100 g, initially at
rest, is hit by a
racket.
The racket is in
contact with the ball
for 20 ms and the
force of contact
varies as shown in the
graph.
What was the speed
of the ball as it left
the racket?
F/N
400
0
20
t / ms
Solution
Impulse
= area under graph
= ½ x (20 x 10-3) x 400
= 4 Ns
Ft = mv – mu
4 = 0.1v – 0
v = 4 / 0.1 = 40 ms-1
2005 Qu: 5
2007
2008
2003 Qu: 5
2004 Qu: 5
Past Papers
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2003 Qu: 22
2006 Qu: 22
2009 Qu: 22(b)
2010 Qu: 23 (a)(ii) & (iii)
Open-ended question
specimen paper
Possible solution
• Hard outer shell absorbs some impact from
the road
• Soft polystyrene foam liner increases contact
time of head
• Using equation F = mv – mu / t it is clear that
to decrease the average force on the head,
contact time should be increased
• Decreasing the average force results in less
damage to the head
Open ended question
2012 Revised Higher
Possible solution
• Compare hitting the ball with and without swinging
‘through the ball’ and using the equation Ft = mv – mu
• In both cases the mass of the ball and the initial
velocity are the same
• If we take u = 0 then Ft = mv
• Assume the racquet used is same in both cases and
applies same force, F
• Not going through the ball means a shorter contact
time, t
• If contact time is small then so is Ft and mv
• As m is a constant then final velocity, v, is small
• Going through increases contact time, t, increases Ft
and as a result, increases v
Questions
• Activity sheets:
• Collisions and explosions
• You should now be able to answer
questions: 14 onwards in class jotter
Answers
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Collisions and Explosions
14. 1·58 ms −1
15. 100 N
16. 3·0 × 10−2 s
17. 2·67 m s −1
18 (a) + 0·39 kg m s −1 if you have
chosen upwards directions to be
positive
(b) + 0·39 N s if you have chosen
upwards directions to be positive
(c) 15·6 N downwards
(d) 15·6 N upwards
(e) 16·6 N upwards
19. (a) v before = 3·96 ms −1
downwards; v after = 2·97 ms −1
upwards
(b) 9·9 × 10 −2 s
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20.(a) Teacher Check
(b) 0·2 s
(c) 20 N upwards (or –20 N for the
sign convention used in the graph)
(d) 4·0 J
21. 1·25 × 103 N towards the wall
22. 9·0 × 104 N
23. Teacher Check
24. (a) (i) 4·0 m s −1 in the direction
the 2·0 kg trolley was travelling
(ii) 4·0 kg m s −1 in the direction the
2·0 kg trolley was travelling
(iii) 4·0 kg m s −1in the opposite
direction the 2·0 kg trolley was
travelling
(b) 8·0 N
25. Teacher Check
26. Teacher Check