1. PHYSICS
Further Mechanics
Andres Igea
Friday 19th October 2012

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

12. Resources
AQA Mechanics 1 book
EDEXCEL Physics A2 Textbook
www.physicsclassroom.com
www.nasa.gov

13. Thanks a lot for
watching
Andres