ABRHS PHYSICS NAME: ________________
Elastic Collisions
side 1
Energy and momentum are always conserved in a collision, no matter what happens. Momentum is
easy to deal with because there is only “one form” of momentum, (p=mv), but you do have to
remember that momentum is a vector. Energy is tricky because it has many forms, the most
troublesome being heat, but also sound and light. If kinetic energy is conserved in a collision, it is
called an elastic collision. In an elastic collision, the total kinetic energy is conserved because the
objects in question “bounce perfectly” like an ideal elastic. An inelastic collision is one where some
of the of the total kinetic energy is transformed into other forms of energy, such as sound and heat.
Any collision in which the shapes of the objects are permanently altered, some kinetic energy is
always lost to this deformation, and the collision is not elastic. It is common to refer to a
“completely inelastic” collision whenever the two objects remain stuck together, but this does not
mean that all the kinetic energy is lost; if the objects are still moving, they will still have some
kinetic energy.
General Equation Derivation: Elastic Collision in One Dimension
Given two objects, m1 and m2, with initial velocities of v1i and v2i, respectively, how fast will they be
going after they undergo a completely elastic collision? We can derive some expressions for v1f and v2f
by using the conservation of kinetic energy and the conservation of momentum, and a lot of algebra.
Begin by making the following conservation statements:
Conservation of Kinetic Energy:
1
2m1
v1i
2+1
2m2v2i
2=1
2m1v1f
2+1
2m2v2f
2
Conservation of Momentum:
m1v1i+m2v2i=m1
v1f+m2v2f
To solve for v1f and v2f (which is really two equations in two unknowns), we need some algebra
tricks to simplify the substitutions. Take both equations and group them according to the masses:
put all the m1’s on one side of the equation and all the m2’s on the other. We’ll also cancel out all
the 1/2’s at this point.
Conservation of Kinetic Energy becomes:
m1v1i
2−m1v1f
2=m2v2f
2−m2v2i
2
which can be simplified as
m1v1i
2−v1f
2
( )
=m2v2f
2−v2i
2
( )
eqn. 1
Conservation of Momentum becomes:
m1v1i−m1v1f=m2v2f−m2v2i
which can be simplified as
m1v1i−v1f
( )
=m2v2f−v2i
( )
eqn. 2
Now comes the algebra fun. Divide equation 1 by equation 2.
m1v1i
2−v1f
2
( )
m1v1i−v1f
( )
=m2v2f
2−v2i
2
( )
m2v2f−v2i
( )
After all the cancellations, we are left with:
v1i+v1f=v2f+v2i
eqn. 3
Solving for v1f we get:
v1f=v2f+v2i−v1i
eqn. 4
Now we take equation 4 and substitute back into one of our original equations to solve for v2f. Since
the momentum equation is easier, lets use that.
Conservation of Momentum becomes:
m1v1i+m2v2i=m1(v2f+v2i−v1i)+m2v2f
