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.
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)v^{2}= 0.025v^{2}. 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: 0.025v^{2}.
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.
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,