Physics 201
Professor P. Q. Hung
311B, Physics Building
Physics 201 – p. 1/2
Linear Momentum and Collisions
Collisions
P~tot,f = P~tot,i : Momentum is always conserved
during a collision.
Physics 201 – p. 2/2
Linear Momentum and Collisions
Collisions
P~tot,f = P~tot,i : Momentum is always conserved
during a collision.
Ktot,f 6= Ktot,i : Inelastic collision.
Physics 201 – p. 2/2
Linear Momentum and Collisions
Collisions
P~tot,f = P~tot,i : Momentum is always conserved
during a collision.
Ktot,f 6= Ktot,i : Inelastic collision.
Ktot,f = Ktot,i : Elastic collision.
Physics 201 – p. 2/2
Linear Momentum and Collisions
Inelastic Collisions
Is our example at the beginning of the class an
inelastic collision?
First we use momentum conservation to
obtain the final speed of of the composite
system of the two objects:
1
V = m1m+m
v1 .
2
Physics 201 – p. 3/2
Linear Momentum and Collisions
Inelastic Collisions
Ktot,i = 21 m1 v12 ;
1 m21
+ m2 )V = 2 m1 +m2 v12 .
m2 2
m2
Ktot,i = − 12 mm11+m
=
−
v
m1 +m2 Ki .
2 1
1
2 (m1
2
Ktot,f =
⇒ Ktot,f −
⇒ Ktot,f < Ktot,i Surely an inelastic collision!
Physics 201 – p. 4/2
Linear Momentum and Collisions
Inelastic Collisions
Ktot,i = 21 m1 v12 ;
1 m21
+ m2 )V = 2 m1 +m2 v12 .
m2 2
m2
Ktot,i = − 12 mm11+m
=
−
v
m1 +m2 Ki .
2 1
1
2 (m1
2
Ktot,f =
⇒ Ktot,f −
⇒ Ktot,f < Ktot,i Surely an inelastic collision!
Some of the initial Kinetic Energy is converted
into heat, for example, during the collision
process, resulting in its loss.
Physics 201 – p. 4/2
Linear Momentum and Collisions
Elastic Collisions
P~tot,i = P~tot,f .
Physics 201 – p. 5/2
Linear Momentum and Collisions
Elastic Collisions
P~tot,i = P~tot,f .
Ktot,i = Ktot,f
Physics 201 – p. 5/2
Linear Momentum and Collisions
Elastic Collisions
One dimension: Take an example of an object
of mass m1 moving to the right with an initial
speed v0 . It collides with an object of mass
m2 , initially at rest. What are the final
velocities?
Physics 201 – p. 6/2
Linear Momentum and Collisions
Elastic Collisions
One dimension: Take an example of an object
of mass m1 moving to the right with an initial
speed v0 . It collides with an object of mass
m2 , initially at rest. What are the final
velocities?
Momentum conservation:
m1 v0 = m1 v1f + m2 v2f (1).
Physics 201 – p. 6/2
Linear Momentum and Collisions
Elastic Collisions
One dimension: Take an example of an object
of mass m1 moving to the right with an initial
speed v0 . It collides with an object of mass
m2 , initially at rest. What are the final
velocities?
Momentum conservation:
m1 v0 = m1 v1f + m2 v2f (1).
Conservation of K.E.:
1
1
1
2
2
2
=
m
v
m
v
m
v
+
2 1 0
2 1 1f
2 2 2f . (2)
Physics 201 – p. 6/2
Linear Momentum and Collisions
Elastic Collisions
(1) and (2) give
m1 −m2
v1f = m
v
0
+m
1
2
1
v
v2f = m2m
0
+m
1
2
Physics 201 – p. 7/2
Linear Momentum and Collisions
Elastic Collisions
Physics 201 – p. 8/2
Linear Momentum and Collisions
Elastic Collisions: Special Cases
m1 = m2 → v1f = 0 and v2f = v0
Physics 201 – p. 9/2
Linear Momentum and Collisions
Elastic Collisions: Special Cases
m1 = m2 → v1f = 0 and v2f = v0
m1 ≪ m2 → v1f ≈ −v0 and v2f ≈
2m1
m2 v0 .
Physics 201 – p. 9/2
Linear Momentum and Collisions
Elastic Collisions: Special Cases
m1 = m2 → v1f = 0 and v2f = v0
m1 ≪ m2 → v1f ≈ −v0 and v2f ≈
2m1
m2 v0 .
m2 ≪ m1 → v1f ≈ v0 and v2f ≈ 2 v0 .
Physics 201 – p. 9/2
Linear Momentum and Collisions
Elastic Collisions
Physics 201 – p. 10/2
Linear Momentum and Collisions
Elastic Collisions
For elastic collision in two dimensions, we have
three equations:
P~tot,i = P~tot,f . (2 equations)
Physics 201 – p. 11/2
Linear Momentum and Collisions
Elastic Collisions
For elastic collision in two dimensions, we have
three equations:
P~tot,i = P~tot,f . (2 equations)
Ktot,i = Ktot,f . (1 equation)
Physics 201 – p. 11/2
Linear Momentum and Collisions
Elastic Collisions
Physics 201 – p. 12/2
Linear Momentum and Collisions
Elastic Collisions
Example: Two metal spheres, suspended by
vertical cords, initially just touch as shown in
class. Sphere 1 with mass m1 = 30g , is pulled to
the left to height h1 = 8.0cm, and then released
from rest. After swinging down, it undergoes an
elastic collision with sphere 2 whose mass is
m2 = 75g. What is the velocity v1f of sphere 1
just after the collision?
Solution:
Calculate v1i of sphere 1√
just before collision.
1
2
m
v
2gh1 = 1.252m/s.
2 1 1i = m1 gh1 → v1i =
Physics 201 – p. 13/2
Linear Momentum and Collisions
Elastic Collisions
1 −m2
From v1f = m
m1 +m2 v0 , one gets
v1f = −0.537m/s. Sphere 1 moves back to
the left.
Physics 201 – p. 14/2
Linear Momentum and Collisions
Center of Mass
The center of mass of a body or a system of
bodies is the point that moves as though all of
the mass were concentrated there and all
external forces we applied there.
Physics 201 – p. 15/2
Linear Momentum and Collisions
Center of Mass
The center of mass of a body or a system of
bodies is the point that moves as though all of
the mass were concentrated there and all
external forces we applied there.
In two dimensions, it will be defined as
+m2 x2 +..
Xcm = m1mx11 +m
(I)
2 +..
+m2 y2 +..
Ycm = m1my11 +m
(II)
2 +..
Physics 201 – p. 15/2
Linear Momentum and Collisions
Center of Mass
From (I,II) it follows that
V~cm =
m1~v1 +m2~v2 +..
m1 +m2 +..
Physics 201 – p. 16/2
Linear Momentum and Collisions
Center of Mass
From (I,II) it follows that
V~cm =
m1~v1 +m2~v2 +..
m1 +m2 +..
~ cm =
A
m1~a1 +m2~a2 +..
m1 +m2 +..
Physics 201 – p. 16/2
Linear Momentum and Collisions
Center of Mass
Physics 201 – p. 17/2
Linear Momentum and Collisions
Center of Mass
P~tot = M V~cm .
Physics 201 – p. 18/2
Linear Momentum and Collisions
Center of Mass
P~tot = M V~cm .
~ cm
F~tot = M A
Physics 201 – p. 18/2
Linear Momentum and Collisions
Center of Mass
Physics 201 – p. 19/2
Linear Momentum and Collisions
Rocket Propulsion
There are situations in which the change in
momentum is due only to the change in the
mass of the object.
Physics 201 – p. 20/2
Linear Momentum and Collisions
Rocket Propulsion
For example the rocket engine ejects fuel with
a speed v. The mass of the ejected fuel is
∆m and its momentum is -(∆m)v (taking the
positive direction to be forward).. The initial
total momentum is zero before the engine is
fired so it remains zero afterwards. Therefore
the momentum change of the rocket is
(∆m)v. The force on the rocket is then
∆m
F = ∆p
=
(
∆t
∆t )v.
Physics 201 – p. 21/2
Linear Momentum and Collisions
Rocket Propulsion
For example the rocket engine ejects fuel with
a speed v. The mass of the ejected fuel is
∆m and its momentum is -(∆m)v (taking the
positive direction to be forward).. The initial
total momentum is zero before the engine is
fired so it remains zero afterwards. Therefore
the momentum change of the rocket is
(∆m)v. The force on the rocket is then
∆m
F = ∆p
=
(
∆t
∆t )v.
Thrust= ( ∆m
∆t )v
Physics 201 – p. 21/2