A bullet of mass m is fired at a block of mass M hanging from a string. The bullet embeds itself in the block. This is an inelastic collision, in contrast to an elastic collision where energy is conserved.
Is the initial momentum of the bullet equal to the momentum of the bullet plus the block immediately after the collision?
Is the initial kinetic energy of the bullet equal to the kinetic energy of the block plus bullet immediately after the collision?
Is the initial kinetic energy of the bullet equal to the gravitational potential energy of the block plus bullet at the top of the swing?
Answer to first question: Yes!! Momentum is always conserved.
Answer to second two questions: No!! The collision is completely
inelastic, resulting in a loss of mechanical energy.
Reminder:
Elastic collision: both
energy and momentum are conserved.
Inelastic collsion:
momentum is conserved, but energy is "lost".
Practical Question: What is v_{o}, the speed of the bullet before it hits the block if the combined block/bullet system swings up a height h?
We solve the ballistic pendulum problem in 2 stages:
1. Energy conservation from just after the collision until the top of the arc.
2. Momentum conservation for the collision.
1. Energy conservation gives:
½ (m+M)v_{f}^{2} = (m+M)gh.
So the speed of the system immediately after the collision is:
v_{f} = (2gh)^{½}.
2. Momentum conservation: to determine the relation between v_{f}
and v_{o}.
Momentum before: mv_{o} =
momentum after: (m+M)v_{f}
→ v_{o}= (2gh)^{½}(M +
m)/m.
How much energy was "lost" in the collision?

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Conservation of momentum, (m+M)v_{f} = mv_{o}, gives:

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