Impulse
Notes p.29
During any collision some momentum is transferred from one object to another.
So… momentum of one object decreases by D p and momentum of the other object increases by
D p.
D p = change in momentum of one object
D p = the IMPULSE of the collision
Note: Impulse is just “change in momentum” so has momentum units!

Now if the velocity of one object, before the collision is “ u
” and the velocity of the same object, after the collision, is “ v
” then impulse = D p = m v
- m u
D p = m ( v – u )

Then consider that during the collision the object experiences an average force , F , over a contact time , t , causing an acceleration , a .
So D p = m(v-u)
= m ( a t)
IMPULSE = D p = F t or m ( v – u) = F t
Note: Impulse units can therefore also be “Ns”.

Impulse Example 1
A 0.8 kg trolley travelling at 0.7 ms -1 hits a wall.
a) The trolley rebounds at 0.3 ms -1 . Calculate the impulse of the force on it. m = 0.8kg
D p = m (v – u) u = 0.7ms
-1
= 0.8 (-0.3 – 0.7) v = -0.3ms
-1
= - 0.8 kg ms -1
D p = ?
Impulse = 0.8 kgms -1 from wall onto trolley

Impulse Example 1 (cont.) b) The trolley is in contact with the wall for 50ms.
Calculate the average force the wall exerts on the trolley during this time.
D p = - 0.8 kgm/s D p = F t t = 50 x 10 -3 s F = D p / t
F = ?
= - 0.8 / 50 x 10 -3
= - 16 N
Average Force is 16N from wall onto trolley

Problems 89 – 91.

Notes p.31
IMPULSE from a Force-Time Graph
A force-time graph can be plotted showing how the force on an object changes during the contact time of a collision.
The equation, m(v-u) = Ft , refers to average
force! In reality, the force on an object would vary like this:
Force (N) time (s)

Force (N)
F time (s)
IMPULSE = F t
= AREA UNDER F-t GRAPH

Impulse Example 2
During a collision the force experienced by a
700kg stationary object was recorded as follows:
800
Force
(N)
0
2 time (s)
Determine the speed of the object after the collision.

800
Force
(N)
0
2 time (s)
1 st IMPULSE = AREA = ½ x 2 x 800 = 800 kgms -1
2 nd
D m = 700 kg u = 0 ms -1 v = ? p = 800kgms -1
D p = m (v – u)
= mv (as u = 0ms -1 ) v = D p / m
= 800 / 700
= 1.14 ms -1

Key Questions on Impulse
A car travelling at 20 ms -1 crashes into a wall and is brought to rest by the wall.
An identical car crashes into the same wall at the same speed, but it has a cushioned bumper so it takes longer to come to rest.
1. In which collision is the impulse greater?
Explain your answer.
Answer
The impulse is the same in each collision!
As the object masses and initial and final velocities are unchanged, then the change in momentum ( D p = m(v – u)) is unchanged.

Key Questions on Impulse
2. Explain why the second car experiences less damage.
Answer
The second car takes longer to come to rest.
The IMPULSE = F t, and the impulse is unchanged.
So, as t has increased, then F must decrease to keep the impulse of the collision the same.
This reduced force means less damage to the car!

3. On the one set of axes, sketch graphs of force against contact time for each collision.
First car – big force, short time.
0
Second car – longer time so smaller force
BOTH SAME
AREA !!
time (s)

The area under the graph is equal to the
.
N.B.
If the impulse stays constant but contact time increases, then the maximum force has to decrease to allow the area under the graph to remain the same. (Think … air bags!)
One practical use of impulse is to find the force exerted on one object by another. For example, you could find the force exerted by a club on a golf ball.

Newton’s Third Law
From National 5 you might remember Newton’s
Third Law …
“Every force has an equal and opposite reaction force.”
This law is the basis of Impulse theory
.
Think outside the box ???!!!
Try to make the connection between Newton’s Third Law and
Impulse theory.

In a collision between 2 objects, each object experiences an equal but opposite force from the other object.
As Impulse = av Force x time then, during a collision, each object must experience equal but opposite impulses.
This means that as one object experiences a
POSITIVE change in momentum, the other will experience the same size but NEGATIVE change in momentum.
So momentum is “transferred” .

Problems to do
Q. 76- 88
76a) 20 kgms -1 , right b) 500kgms -1 , down c) 9kgms -1 , left
77 0.75 ms -1 ,
Ek before = 2.25J, Ek after = 1.125J so INELASTIC
78 2.4 ms -1
81 8.6 ms -1
79 3 kg 80 a) 2.7ms
-1 b) 0.2J
82a) 23ms -1
82 b) INELASTIC … Ek before = 730000J,
Ek after = 702000J .
83 8.67 ms -1 84 0.6ms
-1 in initial direction of 2 nd vehicle
85 16.7 ms -1 86 4kg 87 0.8 ms -1 in opp direction
88 1.3 ms -1

Problems 89 – 99.
89. 1.58 ms
92. 2.67 ms
-1
-1
90. 100N 91. 0.03s
93. a) -0.39 kgms -1 b) 15.6N
94. a) 3.96 ms -1 , 2.97 ms -1 b) 0.1s
95.a)
AB = const acceleration towards wall, BC = in contact with wall slowing down towards wall, CD = in contact with wall speeding up away from wall, DE slowing down to rest moving away from wall. b) 0.2s c) –20N d) 4.0 J
96. 1250N 97. 9x10 4 N 98. See notes
99.a) (i) 4 ms -1 (ii) 4 kgms -1 (iii) 4 kgms -1 b) 8N

EXTRA PRACTICE!
Revision questions for Higher Physics, page 23,
Q. 1 - 16