Motion, Forces and Energy Lecture 8: 1- and 2-d Collisions and Rockets
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Motion, Forces and Energy Lecture 8: 1- and 2-d Collisions and Rockets et’s begin by defining the LINEAR MOMENTUM of a particle of mass m m ith a velocity v as: p = m v We can now redefine Newton’s 2 nd Law as: The rate of change of linear momentum of a part is equal to the net force acting on the particl ma dt dv m dt mv d dt dp F ) (
description
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 = m v. We can now redefine Newton’s 2 nd Law as: The rate of change of linear momentum of a particle - PowerPoint PPT Presentation
Transcript of Motion, Forces and Energy Lecture 8: 1- and 2-d Collisions and Rockets
Motion, Forces and EnergyLecture 8: 1- and 2-d Collisions and Rockets
Let’s begin by defining the LINEAR MOMENTUM of a particle of mass m movingwith a velocity v as:
p = m v
We can now redefine Newton’s 2nd Law as:
The rate of change of linear momentum of a particleis equal to the net force acting on the particle:
madt
dvm
dt
mvd
dt
dpF
)(
Conservation of Momentum For A Two-Particle System
Law of Conservation of Momentum (LCM):
p1i + p2i = p1f + p2f
m2
v2f
v1f
m1
An astronaut in space wants to move. She throwsher jacket in the opposite direction to that in whichshe wishes to move:
ff
ff
ii
vm
mv
LCMvmvm
tatstationaryvmvm
21
21
2211
2211
)(0
)0(0
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.
v1f v2f
v2iv1i
Before collision
After collision
Using Law of Conservation of Momentum andConservation of Kinetic Energy (elastic collision), we can show that:
iif
iif
vmm
mmv
mm
mv
vmm
mv
mm
mmv
221
121
21
12
221
21
21
211
2
2
Case 1: If m1=m2, v1f=v2i & v2f=v1i
(exchange of speeds: pool/snooker).
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.
vf
v2iv1i
Before collision
After collision
iif
fii
vmm
mv
mm
mv
vmmvmvm
221
21
21
1
212211
The Ballistic Pendulum
iffi
fafter
iibefore
vmm
mvsovmmvmLCM
vmmKE
vasvmKE
121
12111
221
22
11
:
2
1
02
1
This is a device used to measure the speed of a fast-moving object such as a bullet.
hm1
v1i
m2
vf
m1+m2
Since KEafter is converted into PE, wecan show that:
hgm
mmv i
2
1
211
An Aside: Rocket Propulsion
vR
vex
Rocket moves upwards due to LCMSince exhaust gases are expelled downwards.
exR
exR
exexRR
vM
Mv
vMvM
0
We can’t write the above since the mass of the
rocket+fuel is continually decreasing!
So we consider a time period t at the beginning of which we say that the mass of the “rocket plus fuel” is MR+m where m is the mass of the emitted fuel gasesin the period t. Also at the start of t, 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:
v2f
m1 m2
v1i
v1fcos
v1f
v1fsin
v2fcos
-v2fsin
Before collision
After collision
Resolving in x and y directions
sinsin0:
coscos:
2211
221111
ff
ffi
vmvmy
vmvmvmx
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.
v2f
m1 m2
v1i
v1fcos
v1fv1fsin
v2fcos
-v2fsin
Before collision
After collision
m1=m2
