Asked by Richard
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
(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
Answers
Answered by
bobpursley
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.
Answered by
Richard
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?)
Answered by
bobpursley
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
Answered by
Richard
Wouldn't it be KEbottom = KEtop without the mgh term since there's no height at the bottom?
Answered by
Richard
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?
Answered by
bobpursley
Yes, it is right if you put the plus sign where you did.
There are no AI answers yet. The ability to request AI answers is coming soon!