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Posted by on Thursday, January 15, 2009 at 11:21pm.

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

  • Physics - URGENT - , Thursday, January 15, 2009 at 11:24pm

    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.

  • Physics - URGENT - , Thursday, January 15, 2009 at 11:31pm

    So mg=mrw^2?

    (and how do I find the w at the top of the loop?)

  • Physics - URGENT - , Thursday, January 15, 2009 at 11:35pm

    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

  • Physics - URGENT - , Thursday, January 15, 2009 at 11:45pm

    Wouldn't it be KEbottom = KEtop without the mgh term since there's no height at the bottom?

  • Physics - URGENT - , Thursday, January 15, 2009 at 11:56pm

    Actually, I think it should be:

    KE of bullet@bottom = KE of block&bullet@top + PE of block&bullet@top

    Is that right?

  • Physics - URGENT - , Friday, January 16, 2009 at 9:31am

    Yes, it is right if you put the plus sign where you did.

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