A small block of mass 5 kg initially rests on a track at the bottom of a circular, vertical loop-the-loop, which has a radius of 0.8 m. The surface contact between the block and the loop is frictionless. A bullet of mass 0.7 kg strikes the block horizontally with initial speed 120 m/s and remains embedded in the block as the block and bullet circle the loop.
(a) Speed of block&bullet immediately after impact
(b) Kinetic Energy of block&bullet when they get midway to the top of the loop
(c) Minimum initial speed Vmin of bullet if block&bullet are to successfully execute a complete circuit of the loop
(a) CONSERVATION OF MOMENTUM
(b) CONSERVATION OF ENERGY
(c) NOT SURE
6 answers
On c, you need to find velocity at the top what will give a centripetal force equal to mg, because if it is less than mg, the block will fall down.
So mg=mrw^2?
(and how do I find the w at the top of the loop?)
(and how do I find the w at the top of the loop?)
mg=mv^2/r
v^2=rg at top, so use that velocity at the top to figure the minimum vleocity at the bottom. (ke at top=keatbottom+mgh
v^2=rg at top, so use that velocity at the top to figure the minimum vleocity at the bottom. (ke at top=keatbottom+mgh
Wouldn't it be KEbottom = KEtop without the mgh term since there's no height at the bottom?
Actually, I think it should be:
KE of bullet@bottom = KE of block&bullet@top + PE of block&bullet@top
Is that right?
KE of bullet@bottom = KE of block&bullet@top + PE of block&bullet@top
Is that right?
Yes, it is right if you put the plus sign where you did.
1 answer