Speeding Bullet

tandoorichicken

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.

faust9

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.

arcnets

Since the collision is completely inelastic, momentum is conserved.

HallsofIvy

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.

arcnets

I think if something hits and sticks, energy is not conserved.

joc

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