# 2 blocks (collision and separation) Simple harmonic motion?

Two blocks of mass m1 and m2 = m1 move together with a constant velocity Vo on a frictionless surface. (clearly, block 2 is much denser than block 1 in this picture). there is no space between the blocks but the blocks are not stuck together.

the blocks move toward a lowered ceiling that is only high enough for block 2 to pass under. there is a horizontal spring of stiffness k attached to the overhang, which block 1 eventually sticks to upon contact.

a)is the total momentum conserved after block 1 collides with the spring? explain.

b)is mechanical energy conserved after block 1 collides with the spring? explain.

c)when block 1 collides with the spring it goes in simple harmonic motion, derive an ex[ression fo the spring constant,k if the spring compresses a distance l before bring block1 to rest.

d)write down the velocity of the center of mass as a function of time.

let t=0 be the time at which block 1 collides with the spring. what is the first time at which Vcm is maximum?

Relevance

a)is the total momentum conserved after block 1 collides with the spring? explain.

NO

As the masses are identical, and the initial velocity is identical, each mass carries one half of the total initial momentum.

As soon as m₁ comes in contact with the spring, some of its kinetic energy is stored in the spring as potential as the mass slows down. At maximum compression, this mass has zero velocity and the total system momentum has been reduced to half of the original. When the spring expands and m₁ velocity becomes - v₀, then total system momentum has been reduced to zero as each half of the original combined mass now has equal and opposite velocity.

b)is mechanical energy conserved after block 1 collides with the spring? explain.

YES

kinetic energy of m₂ remains unchanged

kinetic energy of m₁ is converted to an equal amount of spring potential energy and is returned as kinetic energy when the spring expands again.

c)when block 1 collides with the spring it goes in simple harmonic motion, derive an ex(p)ression fo(r) the spring constant,k if the spring compresses a distance l (L for clarity) before bring block1 to rest.

Initial kinetic energy will completely convert to spring potential energy at maximum compression.

½mv² = ½kx²

in terms of this exercise

½m₁v₀² = ½kL²

k = m₁v₀² / L²

d) write down the velocity of the center of mass as a function of time.

The position of m₁ with the origin at the point of contact at t = 0 can be written as

x = Lcosωt

where ω = √(k/m₁) = √(m₁v₀² / L² / m₁) = √(v₀² / L²) = v₀ / L

the period of this motion is T = 2π√(m₁/k) = 2π√(m₁/(m₁v₀² / L²)) = 2π√(L² / v₀² ) = 2πL / v₀

the velocity of m₁ will be the derivative of position

v₁ = v₀ - ωLsinωt

so

v₁ = v₀ at t = 0

v₁ = 0 at T/4 or t = ½πL / v₀

v₁ = -v₀ at t = T/2 or t = πL / v₀

the velocity of the center of mass of the system is found using the momentum equation

vcm = m₁v₁ + m₂v₀ / (m₁ + m₂)

as m₁ = m₂

vcm = m₁(v₁ + v₀) / 2m₁

vcm = (v₁ + v₀) / 2

vcm = (v₀ - ωLsinωt + v₀) / 2

vcm = (2v₀ - ωLsinωt) / 2

vcm = v₀ - ½ωLsinωt

so

vcm = v₀ at t = 0

vcm = ½v₀ at t = T/4

vcm = 0 for t ≥ T/2

what is the first time at which Vcm is maximum?

t = 0 s

• tazkir4 months agoReport

Thank you so much!

• So how about that picture?

• tazkir4 months agoReport

sorry for some reason it didn't upload.
https://s.yimg.com/tr/i/19814473c78e483391120d0a1605968c_A.png