Perfectly Inelastic Collisions - Prep

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Impulse and Momentum
Conservation of Momentum
Elastic & Inelastic
Study Guide
Momentum is Conserved in an isolated
system.
“Isolated system” means that there are
no external forces acting on the thing.
Collisions can be classified according to
the energy interaction that takes place:
• Elastic collision  kinetic energy is
conserved
• Inelastic collision  kinetic energy is not
conserved
• Perfectly inelastic collision  objects stick
together and have the same velocity.
We begin with a two-body collision where each object has
some velocity prior to the collision. Both objects have
momentum equal to the mass times the velocity.
The total momentum before the collision is:
After the collision the velocities will have changed. The total
momentum will be:
We know that the momentum before must equal the
momentum after
So here is an equation which will work for any two-body collision:
Perfectly Inelastic Collisions: These are collisions where the
two objects stick together after they collide.
The key thing to remember here is that after the collision both
objects stick together and have the same velocity.
After the collision the velocities will have changed.
We know that the momentum before must equal the
momentum after:
Now we can solve for final velocity
Use concept of the conservation of
mechanical energy to solve
this initial velocity of the box right after
The bullet is immersed.
• A 0.05 kg bullet collides and sticks to a 2.5 kg stationary block
suspended from a string. The bullet and block swings to a
maximum height of 12 cm. What was the initial speed of the
bullet?
• A 10.0 gram bullet traveling horizontally strikes and sticks in a
80 gram stationary air track glider. If the combined
bullet/glider moves at 20.0 m/s.
What is the bullet's pre-impact velocity ?
Challenging Problem
An incident ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of negligible
friction shown above. It makes a head-on collision in w1ith at target ball B of mass 0.50 kg at rest
the edge of the table. As a result of the collision, the incident ball rebounds, sliding backwards.
at 0.70 m/s immediately after the collision.
(a) Calculate the speed of the 0.50 kg target ball immediately after the collision.
(b) Calculate the horizontal displacement. (use concept of motion to solve this problem)

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