1. More Common Problems/Situations Involving Work and/or Energy and Its Conservation
Elastic Collisions
Ballistic Pendulum
Compressed Spring
Pendulum

2. Elastic Collisions (both momentum and kinetic energy are conserved)
“Object 1 with mass m1 traveling with velocity v1
collides elastically with object 2 with mass m2 and velocity v2. Find their velocities after the collision.”
See a simulated air track at

3. Example: A 4. 0 kg mass (A) moving with a velocity of 8
Example: A 4.0 kg mass (A) moving with a velocity of 8.0 m/s collides elastically with a 6.0 kg mass (B) moving with a velocity of m/s. Find their velocities after the collision.
Let vA = final velocity of 4.0 kg mass and vB = final velocity of 6.0 kg mass.
We now have two equations and two unknowns that must be solved for vA and vB.

4. The momentum equation is a linear equation in vA and vB, and the kinetic energy equation is a quadratic equation in both vA and vB. You are therefore essentially looking for the intersection of a line and a circle (equal masses) or a line and an ellipse (unequal masses).
These equations may be solved using any one of several methods taught in mathematics classes.
Be aware that although there may be one, two, or no mathematical solutions
to these equations, a mathematical solution may be discarded in light of its physical meaning.

5. The Ballistic Pendulum
A device used to determine the initial velocity of a projectile.
Learn more about the ballistic pendulum at these links:

6. Example: A 32 g “bullet” is fired and embeds itself in a 900 g pendulum bob, and the bullet + bob together swing up to a maximum height of 12 cm. What was the initial speed of the bullet?
Change in PE of (bullet + bob) after the collision
= mgh
= ( )kg * 9.8 m/s/s * 0.12 m
= J
Therefore, the KE of the (bullet + bob) was initially J
= ½mv2 = J
= ½(0.932)v2 = J, so v = 1.54 m/s
We use momentum conservation in the inelastic collision between the bullet and the pendulum bob to find the initial velocity of the bullet.
mbulletvbullet + mbobvbob = m(bullet+bob)v(bullet+bob)
(0.032 kg)(vbullet) + 0 = (0.932 kg)(1.54 m/s)
vbullet = m/s

7. Compressed Spring Fires Ball Into Air
A spring with an elastic constant of 12.0 N/cm is compressed 5.8 cm by a 54 g steel ball. If fired vertically upward, what is its initial velocity after release, and how far above the spring’s equilibrium position does it rise. Assume g = 9.8 m/s/s and ignore the effects of air resistance.
The elastic potential energy of the spring when fully compressed will be converted to kinetic energy and gravitational potential energy of the steel ball.
½ kx2 = ½ mv2 + mgh
½(1200 N/m)(0.058 m)2 = ½(0.054 kg)*v2 + (0.054 kg)(9.8 m/s/s)(0.058 m)
so v = 8.58 m/s
At its maximum height, all this kinetic energy will have been converted to gravitational potential energy.
½ mv2 = mgh
h = v2/2g = (8.58 m/s)2/(19.6 m/s/s)
h = 3.76 m (above equilibrium position)
and h = 3.76 m m = m
(above initial position)

8. The Simple Pendulum
As a pendulum swings, it continuously converts gravitational potential energy to kinetic energy of motion. This periodic motion can be analyzed using LoggerPro or any other video analysis software.
See the following web sites for more information about pendulums:
See for abundant information on work and energy.