A 15.0kg block is attached to a very light horizontal spring of force constant 4

Question

A 15.0kg block is attached to a very light horizontal spring of force constant 450N/m and is resting on a smooth horizontal table. (See the figure below .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00m/s to the right, whereupon the stone rebounds at 2.00m/s horizontally to the left.



a)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.)


x=_________ m

Explanation / Answer

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)^2/450
x=0.36514 m..... This may be correct now!! Cheers!!

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