## Speeding Bullet

A 50-g bullet is shot into the 2-kg can of a ballistic pendulum. The can rises to a height of 1.3m Determine the speed of the bullet just before the collision.

Don't know how to begin.
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 I know you say that you don't know how to begin, but I'm hoping you see that this is a momentum problem. IMO momentum problems are best done by drawing a free body diagrams: one immediatly before an event, one immediatly after (or during) an event and one at the end of the event as felt be the whole system (ie a baseball just prior to being hit by a bat, a baseball while (or just after) being hit by the bat, and the final distance traveled by the baseball). Once you have a good set of FBD's analyze the energies at each point of the overall event. That should get you going.
 Since the collision is completely inelastic, momentum is conserved.

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## Speeding Bullet

Conservation of energy. (You appear to be doing problems from a chapter on "conservation of energy!)

Initially, the bullet with mass 50 g= 0.05 kg has (unknown) speed v. It's kinetic energy is (1/2)(0.05)v2= 0.025v2. Taking the height of the bullet and block at the moment of impact to be 0 potential energy, since the block is not moving, the total energy of bullet and block is the kinetic energy of the bullet: 0.025v2.

The bullet and block together rise to a height 1.3 m above the base height, and have 0 speed there. Their potential energy is (0.05+2)(9.8)(1.3)= 26.117 Joules and is the total energy.

Solve 0.025v2= 26.117.
 I think if something hits and sticks, energy is not conserved.
 you'd need both momentum and energy equations for this question. letting Mb = mass of bullet, M = total mass of can and bullet, Vb = velocity of bullet just before collision, V = velocity of bullet+can immediately after collision, h = height of rise, we can write 2 equations: (momentum) MbVb = MV (energy) (1/2)MV^2 = Mgh solving for Vb, we get Vb = MV/Mb Vb = (M/Mb) sqrt(2gh) Vb = 2.05/0.05 rt(2x9.81x1.3) Vb = 207 ms^-1