2. Further Mechanics
Linear Momentum
Conservation of Linear Momentum
Target check
Elastic and Inelastic Collisions in one
dimension
Target check
Two-dimensional collisions
Applications
Resources
3. Linear Momentum
The linear momentum of a particle of
mass m and velocity v is defined as:
p = m⋅v
The linear momentum is a vector
quantity. Its direction is along v.
4. Conservation of Linear Momentum
p = ∑ p = constant
or:
∑ pi =∑ p f
p1,i + p2,i = p1, f + p2, f
5. Target check
A 2kg marble travels to the right at 0.4 m/s, on
a smooth, level surface. It collides head-on with
a 6kg marble moving to the left at 0.2 m/s.
After the collision, the 2 kg marble rebounds at
0.1 m/s.
Task:
Find the velocity of the 6kg marble after the
collision.
Why are the marbles
so heavy ?
6. Elastic and Inelastic Collisions in
one dimension
Momentum is conserved in any collision,
elastic and inelastic.
Mechanical Energy is only conserved in
elastic collisions.
Perfectly inelastic collision: After colliding, particles coalesce
(stick together). There is a loss of energy.
Elastic collision: Particles bounce off each other without loss
of energy.
Inelastic collision: Particles collide with some loss of energy
(deformation), but don’t coalesce.
7. Perfectly inelastic collision of two
particles
pi = p f Notice that p and v
are are vector
m1v1i + quantities = (m1 + m2 )v f
m2 v2i
and, thus have a
direction (+/-).
Ki − Eloss = K f
1 2 1 2 1 2
m1v1i + m2v2i = ( m1 + m2 )v f + Eloss
2 2 2
There is a loss in energy Eloss
8. Elastic collision of two particles
Momentum is conserved
m1v1i + m2 v2i = m1v1 f + m2 v2 f
Energy is conserved
1 2 1 2 1 2 1 2
m1v1i + m2 v2i = m1v1 f + m2 v2 f
2 2 2 2
9. Target check
A bullet (m = 0.01kg) is fired into a block (0.1 kg) sitting at the edge of a
table. The block (with the embedded bullet) flies off the table (h = 1.2 m)
and lands on the floor 2 m away from the edge of the table.
a.) What was the speed of the bullet?
b.) What was the energy loss in the bullet-block collision?
vb = ?
h = 1.2 m
x=2m
10. Two-dimensional collisions
Two particles:
Conservation of momentum:
pi = p f
m1v1i + m2 v2i = m1v1 f + m2 v2 f
Split into components:
p x ,i = p x , f
m1v1ix + 2 v2 ix =m1v1 fx + 2 v2 fx
m m
p y ,i = p y , f
m1v1iy + 2 v2 iy =m1v1 fy + 2 v2 fy
m m
11. Applications
Conservation of momentum:
Rocket being launched into space.
Rocket gains
momentum in
the upwards
direction
The hot gases
gain
momentum in
the
downwards
direction
Momentum measures the quantity of motion, in simple terms how much stuff is moving (mass) and how fast it is moving (velocity) Look at M1 book, according to Newton it can be considered as a measure of stoppability. So from the clear formula we can deduce that momentum is the product of the mass and the velocity of an object.
In any collision, the total momentum before the collision is equal to the total momentum after the collision, provided that there is no external force acting
To see whether you have understood so far what has been taught, we have a little surprise for you:
The correct terminology for the particles sticking together is the term: coalesce
In this case, the particles will coalesce.
If the collision is elastic, we can also use conservation of energy.
By the law of conservation of momentum , m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Before firing, both the gun and the bullet have zero momentum i.e. m 1 u 1 + m 2 u 2 = 0 After firing, the bullet moves forward, and have a forward momentum So, the gun must have a backward momentum