A 15.0kg block is attached to a very light horizontal spring of force constant 5

Question

A 15.0kg block is attached to a very light horizontal spring of force constant 550 N/m and is resting on a smooth horizontal table. (See the figure below (Figure 1) .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/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.) x= m

Explanation / Answer

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)^2/550N… x=0.33 m

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