# Another Laws of Motion question

1. Sep 21, 2006

### konichiwa2x

Hi this problem stumps me.

Two bricks each having a mass of 0.2 kg tied at the end of a light flexible cord and are passing over a small frinctionless pulley. A 0.1kg block is placed on one of the blocks and removed after 2 seconds. How far will each block move in the first second the 0.1kg block is removed?

My Work:

for one of the blocks: $$0.3(10) - T = 0.3a$$
for the other block: $$T - 2 = 0.2a$$
solving, $$a = m/s^2$$
Then I found the velocities of each block after the 0.1kg block is removed. i,e $$v = 2(2) = 4m/s$$
actually, one of the blocks will have 4m/s and the other -4m/s.

for the block that has -4m/s velocity,
the displacement in the first second:
$$s = -4(1) - \frac{1}{2}(10)(1)$$
$$= -9m$$

and for the other block i got the displacement as 1m.

BOOK SOLUTION: my book says the correct answer is 4m for each block.

Last edited: Sep 21, 2006
2. Sep 21, 2006

### Astronuc

Staff Emeritus
After the one kg block is removed, there is not acceleration of the two blocks. One 2 kg block balances the other, and both contribute to the tension in the cord.

One calculated the speed a 4 m/s for one block and -4 m/s for the other block, i.e. one moves up and the other down. In one second, what is the distance traveled at constant velocity?

3. Sep 21, 2006

### konichiwa2x

allright thanks. got it now

by the way, can you please look at this thread I posted earlier today?
https://www.physicsforums.com/showthread.php?t=132919 It is related to the same topic. I have been practicing laws of motion problems all day and I really want to get it straight. only that question is left..thanks