1. 6. Center of mass, Collisions
6.1. The center of mass
To describe the motion of a system of particles (also for their continuous distribution – a solid body), one introduces the concept of the center of mass.
A baseball bat thrown into the air moves in a
complicated manner but one of its points, the center of mass (CM), moves in a parabolic path.
CM moves as if all the system’s mass were concentrated there and all external forces were applied to it (to be proved).
We define the position of CM for a system of
particles by a vector equation
(6.1)
For a solid body summation in Eq.(6.1) must be replaced by integration
(6.2)
where is a mass density.
(figure from HRW)

2. The center of mass, cont.
For a uniform body (ρ=const), position of its CM depends on the body’s shape only
(6.3)
Examples
Two particles of masses m1 and m2 separated by distance d.
The position of CM we find placing both particles on an x axis.
From Eq.(6.1) one obtains
Putting x1=0, we have
The center of mass of a triangle.
In this case there is no need to use Eq.(6.3), but dividing
a triangle into segments it can be concluded that CM is
placed at the intersection of triangle medians.

3. 6.2. Motion of a system of particles
For a system of n particles one can write from a CM definition (6.1)
(6.4)
Differentiating both sides of Eq.(6.4) vs. time gives
(6.5)
By differentiating Eq.(6.5) in respect to time one obtains
(6.6)
Eq.(6.6) can be rewritten as
(6.6a)
The right side of Eq.(6.6a) includes all forces, both the internal forces between the
particles and the external forces acting from the outside. From the Newton’s third
law the sum of all internal forces reduces to zero and the right side of Eq.(6.6a) is
the vector sum of all external forces, the net force. In this case one obtains
(6.7)
This proves the statement given at the beginning of this chapter.
Neton’s second law applied to the center of mass

4. 6.3. Momentum of a system of particles
Eq.(6.5) can be rewritten as
(6.8)
Thus, the linear momentum of a system of particles (RHS of above equation)
is equal to the product of the total mass of the system and the velocity of the center
of mass:
(6.9)
Calculating the time derivative of Eq.(6.9) we have (6.10)
From above relation and taking into account Eq.(6.7) one obtains
(6.11)
From (6.11) it follows, that if is zero, then
(6.12)
The linear momentum of the system is changed by the net external force (Newton’s second law for a system of particles).
The momentum of a system is not changed if it is isolated
(law of momentum conservation for a system of particles).

5. 6.4. Collisions
The object’s momentum can be changed by the collision with other object.
The forces during collision act for a short time and are large. An example is the
collision of a baseball with a baseball bat.
From the Newton’s second law
(6.13)
The net change in the momentum is obtained after integrating Eq.(6.13) in time Δt
or (6.14)
The change in objet’s mometum is equal to the force impuls.
(figure from HRW)
In the figure to the left we see the magnitude of force
varying with time for the collision of a ball with a bat.
The area under the curve is equal to the force impulse in
the collision.

6. Collisions, cont.
In many cases we consider the system isolated in which the objects mutually
collide. In that case the vector of linear momentum of the system cannot change.
The total kinetic energy of colliding bodies can be conserved and in this case
we have an elastic collision.
When the kinetic energy of the system is not conserved, a collision is called
inelastic. If the colliding bodies stick together, the collision is called completely
inelastic.
Two bodies collide moving along an x axis. From the conservation of a linear
momentum
(6.15)
The total kinetic energy is also conserved
(6.16)
Elastic collision in one dimension

7. Collisions, cont.
Solving the simulaneous equations (6.15) and (6.16) for u1 and u2 one obtains
When the particles have the same masses,
they exchange the velocities after collision
Completely inelastic collision in one dimension
From the conservation of linear momentum
we have
(6.17)
and hence the velocity of sticked together masses is
It is easy to check that the kinetic energy of the body after the collision is less than the sum of kinetic energies of the bodies before the collision.

8. Collisions, cont. Sample problem
In figure below mass m2 is at rest on a frictionless surface and touches the end of an unstretched spring of spring constant k. The other end of the spring is fixed to the wall. Mass m1 travelling at speed v1 collides with mass m2 and the two masses stick together. When the masses momentarily stop, by what distance is the spring compressed?
From the conservation of linear
momentum we have
u0 – velocity at the beginning of compression
The kinetic energy of a system of masses
tranfers to the potential energy of a spring