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

Question

A 15.0 kg block is attached to a very light horizontal spring of force constant 450 N/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 8.00 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.( Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

Explanation / Answer

before collision

m1 = 15 kg             m2 = 3 kg

u1 = 0               u2 = 8 m/s

initial momentum Pi = m1*u1 + m2*u2

after collision

v1 = ?                     v2 = ?

final momentum Pf = m1*v1 + m2*v2

from momentum conservation

Pi = Pf

m1*u1 + m2*u2 = m1*v1 + m2*v2

3*8 = 15*v1 + 3*v2.........(1)

from energy conservation

KEi = KEf

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

3*8^2 = 15*v1^2 + 3*v2^2....(2)

solving 1 & 2

v1 = 2.67 m/s

v2 = -5.33 m/s

after the block compresses the spring by x

KE of the block after collision = PE stored in spring

0.5*m1*v1^2 = 0.5*k*x^2

0.5*15*2.67^2 = 0.5*450*x^2

x = 0.487 m <--------------answer

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