Impulse
Mechanics & Properties of Matter 7:
Impulse
AIM
This unit introduces the useful idea of impulse and then defines it in terms of momentum.
Calculations involving impulse and change of momentum are tackled.
OBJECTIVES
On completing this unit you should be able to:
• state that impulse is the product of average force and time of contact.
• state that impulse equals the change in momentum of a body.
• carry out calculations using the following relationships:
o impulse = Ft
o impulse = ∆mv
o Ft = ∆mv
• apply the law of conservation of momentum to the interaction of two objects moving
in one dimension to show that
o the changes in momentum of each object are equal in size and opposite in
direction.
o the forces acting on each object during the interaction are equal in size and
opposite in direction.
Strathaven Academy
-1-
Mechanics and Properties of Matter
Impulse
Impulse
An object is accelerated by a force F for a time, t. The unbalanced force is given by:
F = ma =
Unbalanced force =
m(v - u)
mv - mu
=
t
t
change in momentum
= rate of change of momentum
time
Ft = mv - mu
The term Ft is called the impulse and is equal to the change in momentum.
Note: the unit of impulse, Ns will be equivalent to kg ms-1.
The concept of impulse is useful in situations where the force is not constant and acts for a
very short period of time. One 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 as shown below.
F
t
Since F is not constant the impulse (Ft) is equal to the area under the graph. In any
calculation 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 that for higher physics we approximate this graph to a triangle to allow the calculation of
the area using ½bh.
Examples
1. In a snooker game, the cue ball, of mass 0.2 kg, is accelerated from the rest to a velocity
of 2 ms-1 by a force from the cue which lasts 50 ms. What size of force is exerted by the
cue?
u = 0 v = 2 ms-1 t = 50 ms = 0.05 s m = 0.2 kg
F=?
Ft = mv - mu
F × 0.05 = 0.2 × 2
F=8N
Strathaven Academy
-2-
Mechanics and Properties of Matter
Impulse
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 over this period as shown in the
graph. Determine the speed of the ball as it leaves the racket.
F /N
400
20
Impulse = Area under graph
= ½ × 20 × 10-3 × 400 = 4 N s
u = 0 m = 100 g = 0.1 kg v = ?
Ft = mv - mu = 0.1v
4 = 0.1 v
v = 40 ms-1
t /m s
1
2
3. A tennis ball of mass 0.1 kg travelling horizontally at 10 ms-1 is struck in the opposite
direction by a tennis racket. The tennis ball rebounds horizontally at 15 ms-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 ms-1 v = -15 ms-1 (opposite direction to u)
t = 50 ms = 0.05 s
Ft = mv - mu
0.05 F = 0.1 × (-15) - 0.1 × 10
= -1.5 - 1 = -2.5
- 2.5
F =
= -50 N (Negative indicates force in opposite direction to initial velocity)
0.05
Strathaven Academy
-3-
Mechanics and Properties of Matter
Impulse
ACTIVITY 14
Title: Impulse
Aim: To calculate average force exerted by a “putter” on a golf ball.
Apparatus: 1 “putter” and mountings, 1 metal painted golf ball with metal “tee,” 1 light
Apparatus: gate, 2 millisecond timers, scales.
P iv o t
M illis e c o n d
M illis e c o n d
t im e r
t im e r
L ig h t
gate
P u tter
2
0 .0 0 0
1
head
0 .0 0 0
S w in g
M etal
"tee"
Instructions
Measure the mass of the golf ball, m, and the diameter of the golf ball.
Reset both millisecond timers to zero.
Place the light gate in front of the ball so that it will pass through the centre of the beam after
having been struck by the “putter” head.
Pull the “putter” head back and allow it to strike the ball.
Note the time of contact on timer 1.
Using the time recorded on millisecond timer 2, calculate the velocity, v, of the golf ball after
being struck by the “putter” head.
Use the equation Ft = mv - mu to calculate the force acting on the ball.
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-4-
Mechanics and Properties of Matter
Impulse
NEWTON’S THIRD LAW OF MOTION
We will apply the law of conservation of momentum to a particular interaction between two
objects – an explosion.
va
ma
BEFORE
Now,
AFTER
total momentum = 0
total momentum before
So,
0
Giving,
vb
mb
-(mv)a
total momentum = (mv)a + (mv)b
=
total momentum after
=
(mv)a + (mv)b
=
(mv)b
Since the original momentum of each object was zero, these also represent the changes in
momentum.
So,
(a) The changes of momentum of each object are equal in size
and opposite in direction.
The time for the change in momentum is the same for each object.
So we can write,
-(mv)a
t
=
(mv)b
t
And since force is change in momentum ,
time
We have
So,
Conclusion:
-Fa
=
Fb
(b) The forces acting on each object during the interaction
are equal in size and opposite in direction.
Newton’s third law can therefore be stated in terms of
changes in momentum or in terms of forces.
Strathaven Academy
-5-
Mechanics and Properties of Matter
Impulse
Impulse
89.
A cue exerts an average force of 7 N on a stationary snooker ball of mass 200 g. If the
impact lasts for 45 ms, with what speed does the ball leave the cue?
90.
A girl kicks a football of mass 500 g which was originally stationary. Her foot is in
contact with the ball for a time of 50 ms and the ball moves off with a speed of
10 ms-1. Calculate the average force exerted on the ball by her foot.
91.
A stationary golf ball is struck by a club. The ball which has a mass of 100 g moves
off with a speed of 30 ms-1. If the average force of contact is 100 N calculate the time
of contact.
92.
The graph shows how the force exerted on a hockey ball by a hockey stick varies with
time. If the mass of the ball is 150 g determine the speed of the ball as it leaves the
stick (assume that it was stationary to begin with).
F/N
40 N
20
93.
t / ms
A ball of mass 100 g falls from a height of 20 cm onto a surface and rebounds to a
height of 18 cm. The duration of impact is 25 ms. Calculate:
(a) the change in momentum of the ball caused by the ‘bounce’
(b) the average force exerted on the ball by the surface.
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Mechanics and Properties of Matter
Impulse
94.
A rubber ball of mass 40 g is dropped from a height of 0.8 m onto the pavement.
It rebounds to a maximum height of 0.45 m. The average force of contact between the
pavement and the ball is 2.8 N.
(a) Calculate the velocity of the ball just before it hits the ground and the velocity just
after hitting the ground.
(b) Calculate the time of contact between the ball and pavement.
95.
A ball of mass 400 g travels horizontally along the ground and collides with a wall.
The velocity / time graph below represents the motion of the ball for the first 1.2
seconds.
v / m s-1
A
-4
(a)
(b)
(c)
(d)
B
6
C
0.6
0.8
1.2
t/s
E
D
Describe the motion of the ball during sections AB, BC, CD and DE.
What is the time of contact with the wall?
Calculate the average force between the ball and the wall.
How much energy is lost due to contact with the wall?
96.
Water is ejected from a fire hose at a rate of 25 kg s-1 and a speed of 50 ms-1. If the
water hits a wall calculate the average force exerted on the wall. Assume that the
water does not rebound from the wall.
97.
A rocket burns fuel at a rate of 50 kg per second, ejecting it with a constant speed of
1800 ms-1. Calculate the force exerted on the rocket.
98.
Describe in detail an experiment which you would do to determine the average force
between a football boot and a football as it is being kicked. Draw a diagram of the
apparatus and include all measurements taken and calculations carried out.
99.
A 2 kg trolley travelling at 6 ms-1 collides with a stationary 1 kg trolley.
(a) If they remain connected, calculate:
(i) their combined velocity
(ii) the momentum gained by the 1 kg trolley
(iii) the momentum lost by the 2 kg trolley.
(b) If the collision time is 0.5 s, find the force acting on each trolley.
Strathaven Academy
-7-
Mechanics and Properties of Matter