# The figure below shows block 1 of mass 0.200 kg sliding to the right over a frictionless elevated surface at a speed of 5.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.160 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 = 3.60 m. What is the value of d?

## To find the value of d, we need to consider the conservation of energy.

Let's start by calculating the potential energy (PE) of block 1 when it is at the top of the elevated surface. The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

PE1 = m * g * h
= 0.200 kg * 9.8 m/s^2 * 3.60 m
= 7.056 J

Since block 1 undergoes an elastic collision with block 2, the total kinetic energy (KE) before the collision must be the same as the total kinetic energy after the collision.

The initial KE of block 1 is given by KE1 = (1/2) * m * v^2, where m is the mass of block 1 and v is its velocity.

KE1 = (1/2) * 0.200 kg * (5.00 m/s)^2
= 2.5 J

After the collision, block 2 oscillates in SHM, so its total energy is the sum of its potential energy and kinetic energy. The formula for the period of oscillation is T = 2π * sqrt(m/k), where m is the mass of block 2 and k is the spring constant.

T = 0.160 s
k = 1208.5 N/m

Solving the formula for T, we can find the total energy (E) of block 2:

E = (1/2) * m * (2π/T)^2 + (1/2) * k * x^2

Since block 2 is in SHM, when it reaches the maximum displacement x (amplitude), all of its energy is potential energy, and when it is at the equilibrium position, all of its energy is kinetic energy. Thus, E = max potential energy = (1/2) * k * x^2.

Since we know T, we can use it to find the amplitude x:

T = 2π * sqrt(m/k)
0.160 s = 2π * sqrt(m/1208.5 N/m)

Solving for sqrt(m/1208.5 N/m) gives:

sqrt(m/1208.5 N/m) = 0.160 s / 2π
m/1208.5 N/m = (0.160 s / 2π)^2
m/1208.5 N/m = 0.00639 s^2
m = 1208.5 N/m * 0.00639 s^2

Now, we can find x by substituting m into the equation:

x = sqrt(2 * max potential energy / k) = sqrt(2E/k)
= sqrt(2 * (1/2) * k * x^2 / k)
= sqrt(x^2)
= x

Therefore, the amplitude x is equal to the displacement x, which is equal to the maximum displacement x.

Now, let's find the maximum potential energy of block 2:

E = (1/2) * k * x^2
= (1/2) * 1208.5 N/m * (x^2)

Since the total energy remains constant, the maximum potential energy E is equal to the initial kinetic energy of block 1:

E = KE1 = 2.5 J

Therefore,

2.5 J = (1/2) * 1208.5 N/m * (x^2)
x^2 = (2.5 J * 2) / (1208.5 N/m)
x^2 = 0.00412 m^2
x = sqrt(0.00412 m^2)
x = 0.0642 m

Now, let's find the displacement d of block 1. Since all the energy of block 1 was converted to potential energy, we can equate the initial potential energy of block 1 (PE1) to the final potential energy of block 1 (PE1').

PE1 = PE1' = m * g * d

7.056 J = 0.200 kg * 9.8 m/s^2 * d
d = 7.056 J / (0.200 kg * 9.8 m/s^2)
d = 3.6 m

Therefore, the value of d is 3.6 meters.