# Need Physics Help !!?

A 15.0kg block is attached to a very light horizontal spring of force constant 500 N/m and is resting on a smooth horizontal table. (See the figure below .) Suddenly it is struck by a 3.00kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00m/s horizontally to the left.

Find the maximum distance that the block will compress the spring after the collision.(Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

### 3 Answers

- kuiperbelt2003Lv 71 decade agoBest Answer
there is no figure so I have to guess that the stone's motion lies in the same direction as the spring; if this is not correct then my solution will not be valid

first find the speed of the block immediately after collision; momentum conservation tells us

momentum before collision = momentum after collision

call the initial momentum of the stone as the positive direction, so we have

momentum before = 3kgx 8m/s = 24kgm/s

after collision = 15V + 3kg(-2m/s/s)=15V-6kgm/s

V=speed of block after collision; remember that momentum is a vector so you have to use a negative sign if the motion changes direction

so we have

24 = 15V - 6

V=2m/s

now use conservation of energy to find the compression of the spring

1/2 mv^2 = 1/2 kx^2

x=sqrt[mv^2/k]=sqrt[15kgx(2m/s)^2/500Nm]

x=0.35 m

- goodsonLv 43 years ago
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