A 15.0 kg block is attached to a very light horizontal spring of force constant
ID: 1881484 • Letter: A
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 (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.
Part 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.)
0.390.39
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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 (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.
Part 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 =0.390.39
mSubmitPrevious AnswersRequest Answer
Incorrect; Try Again; 3 attempts remaining
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Figure1 of 1
15.0 kg 3.00 kg 8.00 msExplanation / Answer
Given,
Mass, m1 = 15 kg, m2 = 3 kg
Velocity, V2f = - 2 m/s
Spring constant, k = 450 N/m
Initial velocity, v2i = 8 m/s
Applying law of conservation of momentum
m1 v1i + m2 v2i = m1v1f + m2 v2f
15 x 0 + 3 x 8 = 15 v1f + 3(-2)
V if = 2 m/s
0.5 kx^2 = 0.5 mv^2
450 x x^2 = 15 x 2^2
X = 0.365 m
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