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## Homework Statement

A block with mass m which is on a slippery floor is attached to a spring with spring constant k.

Another block which mass is also m approach and then hit the first block with velocity v as shown in the figure.

After the collision, both blocks stick together and together oscillates on the spring.

The amplitude of the oscillation is

A. A = v √(m/k)

B. A = v √(k/m)

C. A = v √(2m/k)

D. A = v √(m/2k)

E. A = 2v √(m/k)

## Homework Equations

Momentum and energy conservation (I think)

## The Attempt at a Solution

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Since both blocks stick together, it's inelastic collision.

m v = 2m v'

v' = (1/2)v

1/2 (2m) (v')^2 = 1/2 k (Δx)^2

m (1/4) v^2 = 1/2 k (Δx)^2

Δx^2 = (1/2)v^2m/k

Δx = v √(m/2k)

So, I choose D as the answer..

But, the book says that it is C

which is very confusing how to get v = (2π/T) * A.. The explanation is very bad..

Please help