1.3c Impulse
Obvious question: you’re stranded by a fire on an upper floor and have to jump.
Would you rather jump onto grass or concrete? Why? Does what you land on
affect your speed of fall? What difference does a different material make when
you land on it? Try to use precise physics terminology rather than everyday
language.
What is the physics behind this situation? Let’s look at some numbers.
A student jumps from a window ledge 2 m high. Find their velocity when they
reach the ground. This will be the same whatever surface they land on.
When the student lands, the different surfaces have a different amount of ‘give’.
This means the time for their deceleration will be different. For this example we
will take the time of landing on the concrete to be 0.01 s and on the grass to be
0.3 s, as the grass gives way underneath.
Estimate the average force exerted on the student by the concrete and grass.
(b)
Find the deceleration and therefore the average force:
concrete
grass
What do you think about these results?
If you had no choice but to jump onto concrete, what could you do to help
minimise the chance of injury?
In each of these examples, the change in momentum is the same since the speed
at which the student hits the ground is the same. The di fference is the time in
which the change in momentum takes place. Rearranging Newton’s second law
helps to make this more explicit.
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Momentum and Newton’s second law
Newton’s second law: F  ma
but
so
a
vu
t
 v  u  mv  mu
F  m

t
 t 
which is the rate of change of momentum. This how Newton himself defined his
second law.
The law can be rearranged:
Ft  mv  mu
The product Ft is called the impulse. Impulse is a vector quantity.
The units of Ft could be kg m s –1 since these are the units of momentum, but they
are also Ns, and these are the units that should be used for impulse.
The force calculated from the impulse relationship is the average force. The
force involved is rarely constant, eg consider a te nnis ball hit by a racquet:
The force exerted by the racquet on the ball will change with time. The impulse
is calculated from the area under the graph, so:
impulse = area under a force–time graph
This is the case regardless of the shape of the graph. Leading on from this, what
else is represented by the area under a force –time graph?
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Worked examples
1.
In a snooker game, the cue ball, of mass 0.2 kg, is accelerated from the rest
to a velocity of 2 m s –1 by a force from the cue which lasts 50 ms. what
size of force is exerted by the cue?
u = 0, v = 2 m s –1 , t = 50 ms = 0.05 s, m = 0.2 kg, F = ?
Ft  mv  mu
F × 0.05 = (0.2 × 2) – (0.2 × 0)
F × 0.05 = 0.4
F=8N
2.
A tennis ball of mass 100 g, initially at rest , is hit by a racquet. The racquet
is in contact with the ball for 20 ms and the force of contact varies over
this period, as shown in the graph. Determine the speed of the ball as it
leaves the racquet.
impulse
= area under graph
= ½ × 20 × 10 –3 × 400 = 4 N s
u=0
m = 100 g = 0.1 kg
v=?
Ft  mv  mu
4 = 0.1v – (0.1 × 0)
4 = 0.1v
v = 40 m s –1
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3.
A tennis ball of mass 0.1 kg travelling horizontally at 10 m s –1 is struck in
the opposite direction by a tennis racket. The tennis ball rebounds
horizontally at 15 m s –1 and is in contact with the racket for 50 ms.
Calculate the force exerted on the ball by the racket.
m = 0.1 kg
u = 10 m s –1
v = -15 m s –1 (opposite direction to u)
t = 50 ms = 0.05 s
Ft  mv  mu
= (0.1 × (–15)) – (0.1 × 10)
= –1.5 – 1
= –2.5
F = –50 N (the negative sign indicates force in opposite direction to
the initial velocity)
0.05F
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