Question
Asked Nov 25, 2019
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The figure shows block 1 of mass 0.155 kg sliding to the right over a frictionless elevated surface at a speed of 7.65 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1298 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h = 4.60 m. What is the value of d?

2 k
ROROR
h
- d--
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2 k ROROR h - d--

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Expert Answer

Step 1

Since the spring does not affect the collision, we can use the elastic collision equations.

The mass of block 2 can be calculates from the period of SHM.

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m2 Т - 2л, k кт? т, 4л? (1298 N/m)(0.140s) 4л? =0.645kg

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Step 2

At this point, the rebound speed of bloc...

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т - т, т + т, о.155kg - 0.645kg 0.155kg 0.645kg 7.65m/s 4.68 m/s

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