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A 7.90g bullet is travelling at 200ms−1. It hits a 3.00kg sack of sand which is hanging by a rope from the ceiling. The bullet goes into the sack, and is stopped inside it by friction with the sand. How fast is the sack going immediately after the bullet has 'stopped' inside it ('stopped' means stopped relative to the sand, not stopped relative to a stationary observer)? NB you must give your answer to 3 significant figures to be awarded the mark.
Original post by luke.whitlock
A 7.90g bullet is travelling at 200ms−1. It hits a 3.00kg sack of sand which is hanging by a rope from the ceiling. The bullet goes into the sack, and is stopped inside it by friction with the sand. How fast is the sack going immediately after the bullet has 'stopped' inside it ('stopped' means stopped relative to the sand, not stopped relative to a stationary observer)? NB you must give your answer to 3 significant figures to be awarded the mark.


What have you done or tried?

You need the principle of conservation.
Original post by Eimmanuel
What have you done or tried?

You need the principle of conservation.

mvbefore = mvafter
i did (0.0079*200)+(3*0) but i dont understand how to get the second part of the equation?
Original post by luke.whitlock
mvbefore = mvafter
i did (0.0079*200)+(3*0) but i dont understand how to get the second part of the equation?


What do you mean by 2nd part of the equation?
Original post by Eimmanuel
What do you mean by 2nd part of the equation?

well ive done only done the letf side of the equation, i havent done the right side of the equal sign to be able to rearrange
Original post by luke.whitlock
well ive done only done the letf side of the equation, i havent done the right side of the equal sign to be able to rearrange


The bullet and the sack of sand would move with a common velocity after the inelastic collision.
Reply 6
Original post by luke.whitlock
A 7.90g bullet is travelling at 200ms−1. It hits a 3.00kg sack of sand which is hanging by a rope from the ceiling. The bullet goes into the sack, and is stopped inside it by friction with the sand. How fast is the sack going immediately after the bullet has 'stopped' inside it ('stopped' means stopped relative to the sand, not stopped relative to a stationary observer)? NB you must give your answer to 3 significant figures to be awarded the mark.

Use the formula for momentum, p=mv to find initial momentum. Then divide this by the mass of the bullet and the sand bag to find velocity, which is 0.525 ms^-1 to 3 sig figs :smile:

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