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)
Momentum and energy conservation (I think)
The Attempt at a Solution
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..