Download Document

Motion, Forces and Energy
Lecture 8: 1- and 2-d Collisions and Rockets
Let’s begin by defining the LINEAR MOMENTUM of a particle of mass m moving
with a velocity v as:
p=mv
We can now redefine Newton’s 2nd Law as:
The rate of change of linear momentum of a particle
is equal to the net force acting on the particle:
dp d (mv)
F 

dt
dt
dv
m
 ma
dt
Conservation of Momentum For A Two-Particle System
Law of Conservation of Momentum (LCM):
p1i + p2i = p1f + p2f
An astronaut in space wants to move. She throws
her jacket in the opposite direction to that in which
she wishes to move:
v1f
m1v1i  m2 v2 i  0 ( stationary at t  0)
m1
m1v1 f  m2 v2 f  0 ( LCM )
v2f
m2
 m2 
v1 f     v2 f
 m1 
Elastic and Inelastic Collisions in One Dimension
Momentum is conserved in any collision in which external forces are negligible.
However, kinetic energy may (elastic) or may not (inelastic) be conserved.
v1i
v2i
Before collision
v1f
v2f
After collision
Using Law of Conservation of Momentum
andConservation of Kinetic Energy (elastic
collision), we can show that:
Case 1: If m1=m2, v1f=v2i & v2f=v1i
(exchange of speeds: pool/snooker).
 m1  m2 
 2m2 
 v1i  
 v2i
v1 f  
 m1  m2 
 m1  m2 
 2m1 
 m2  m1 
 v1i  
 v2i
v2 f  
 m1  m2 
 m1  m2 
Case 2: If v2i=0 (initially) then (a) if m1>>m2, v1f~v1i and v2f~2v1i &
(b) if m2>>m1, v1f~-v1i and v2f~v2i=0
Perfectly Inelastic Collisions
These are collisions in which the colliding objects actually stick together (merge)
during the collision; thus after the collision, there is only one mass (m1+m2)
involved.
v1i
v2i
Before collision
vf
After collision
m1v1i  m2 v2i  m1  m2  v f
 m1 
 v1i
v f  
 m1  m2 
 m2 
 v2i
 
 m1  m2 
The Ballistic Pendulum
This is a device used to measure the speed of a fast-moving
object such as a bullet.
1
2
KEbefore  m1v1i as v2i  0
2
1
2
KEafter  m1  m2  v f
2
v1i
m1
m1+m2
m2
vf
Dh
LCM : m1v1i  m1  m2  v f
 m1 
 v1i
so v f  
 m1  m2 
Since KEafter is converted into PE, we
can show that:
 m1  m2 
 2 gDh
v1i  
 m1 
An Aside: Rocket Propulsion
vR
Rocket moves upwards due to LCM
Since exhaust gases are expelled
downwards.
M R vR  M ex vex  0
vex
 M ex 
 vex
vR  
 MR 
We can’t write the above since the mass of the
rocket+fuel is continually decreasing!
So we consider a time period Dt at the beginning of which we say that the mass of
the “rocket plus fuel” is MR+Dm where Dm is the mass of the emitted fuel gases
in the period Dt. Also at the start of Dt, the velocity of the rocket is VR.
Two-dimensional Collisions
The physics behind collisions in two dimensions is the same as we have seen
already, except that here we must resolve into x and y components of momentum:
v1fsinq
v1fcosq
Before collision
v1i
m1
v1f
q
f
m2
After collision
v2fcosq
-v2fsinq
v2f
Resolving in x and y directions
x : m1v1i  m1v1 f cosq  m2v2 f cos f
y : 0  m1v1 f sin q  m2v2 f sin f
Example: A pool ball moving at 3.4 ms-1 strikes a stationary ball such
that the first ball continues at a speed of 2.5 ms-1 at an angle of 30o to
its original direction. Ignoring friction, rotational motion and assuming
the collision is perfectly elastic, find the velocity of the struck ball and
its direction.
v1fsinq v
1f
v1fcosq
Before collision
v1i
m1
q
m2
f
After collision
m1=m2
-v2fsinq
v2fcosq
v2f