Momentum
Momentum/linear momentum is defined as
the product of mass and velocity
p = m.v
(kg.m/s) = (kg)(m/s)
Within a closed system, conservation of momentum applies
vA +
A
vA -
vB -
B
vB +
mA vA + mB vB = mA v’A + mB v’B
NOTE!!
The direction of the velocity matters, so please take attention
whether it goes to the left, right, up or down
F.Δt = Δp
F.Δt = m(v-u)
F=
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∆�
A
B
It is considered to take the
right direction as positive
Conservation of Momentum
●
We have to consider objects
which form a closed system –
that is, no external force acts on
them
Conservation of momentum
● Within a closed system, the total
momentum in any direction is constant.
For a closed system, in any direction:
total momentum of objects before collision = total momentum of objects after collision
Momentum - Modelling Collision
●
Springy Collision
- Usually happen in Head-on collision
- The afterward object is not being integrated
●
Sticky Collision
- The afterward objects are sticked together (masses are added)
Momentum
Elastic collision happens when Kinetic Energy is conserved
Total intial kinetic energy = total kinetic energy after collision
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Relative speed of approach and
separation
means speed of one object compared to
another object
For Elastic collision
relative speed of approach = relative
speed of separation
vA
vB
A
B
vA
vB
A
B
Worked Example
In the game of bowls, a player rolls a large ball
towards a smaller, stationary ball. A large ball of
mass 5.0kg moving at 10.0ms−1 strikes a
stationary ball of mass 1.0kg. The smaller ball
flies off at 10.0ms−1.
a Determine the final velocity of the large ball
after
the impact.
b Calculate the kinetic energy ‘lost’ in the impact.
Answer:
total momentum before collision
= total momentum after collision
(5.0 × 10) + (1.0 × 0) = (5.0 × v) +
(1.0 × 10)
50 + 0 = 5.0v + 10
v = 40/5.0 = 8.0 ms−1
1
total k.e. before collision = 2 ×
5.0 × 102 + 0 = 250 J
1
total k.e. after collision = 2 ×
1
5.0 × 8.02 + 2 × 1.0 × 102
= 210 J
k.e. ‘lost’ in the collision =
250 J − 210 J = 40 J
Worked Example
Show that
momentum is conserved in this collision
Y-direction
Before collision: p = 0
(because particle 1 is moving in the x-direction and
particle 2 is stationary).
After collision:
p1= 3.0cos36.9° ≈ 2.40kgms−1 upwards
p2= 4.0cos53.1° ≈ 2.40kgms−1 downwards
These components are equal and opposite and hence
their sum is zero. Hence momentum is conserved in
the
y-direction
X-direction
Before collision: momentum = 5.0 kgms−1 to
the right
After collision:
p1 = 3.0 cos53.1° ≈ 1.80kgms−1 to the right
p2= 4.0 cos36.9° ≈ 3.20 kgms−1 to the right
total momentum to the right = 5.0 kgms−1
Hence momentum is conserved in the xdirection