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 300 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/shorizontally 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.)

Explanation / Answer

here,

mass of stone, m1 = 3kg
velocity of stone, v1 = 8 m/s
vleocity after collision of stone = -2.0 m/s

mass of block, m2 = 15 kg
v2 and v2' are velocity of black before an after collision.

spring constant = 300 N/m

From conservation of Momentum :
initial momentum = Final momentum
m1v1 + m2v2 = m1v1' + m2v2'
m1v1 - m1v1' = m2v2'

solving for v2'

v2' = (m1v1 - m1v1')/m2
v2' = (3*8 + 2*3)/15
v2' = 2 m/s

From Conservation of energy we have :

Kinetic energy gained by block = potential Energy of Spring

0.5*m2*v2'^2 = 0.5 * k * x^2

solving for max compression, x

x = sqrt(m2*v2'^2/k)
x = sqrt(15*2^2/300)
x = 0.447 m

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