IIT JEE , AIEEE Forum - Physics , Mechanics - centre of mass,linear momentum and collision

kislay

pls solve

yahiyafirdous

Hi James Bond

Observe the solution:

At any time t, position of c.m. is v₀t

Hence:

v₀t= (m₁x₁+m₂x₂)/(m₁+m₂)

Given x₁=v₀t - A(1-coswt)

x₂= v₀t +(m₁/m₂)A(1-coswt) ........Ans

x₂-x₁=l₀+x.............................(1)

dx₁/dt= -Awsinwt + v₀..............(2)

dx₂/dt = (m₁/m₂)Aw sinwt + v_0......(3)

Now we take relative velocity of each particle w.r.t. their c.m.

dx'₁/dt= dx₁/dt - v₀= -Awsinwt

dx'₂/dt= dx₂/dt - v₀ = (m₁/m₂)Aw sinwt

d(x₂' - x'₁)/dt=dx/dt=[(m₁+m₂)/m₂]Awsinwt .....(4)

Now equating total P.E. of spring with total energy at extreme position of above oscillation:

(1/2)kl₀²= (1/2) k[(m₁+m₂)/m₂]² A²

l₀=[(m₁+m₂)/m₂]A Ans

rajat

is this the answer

x2 = Vot + m1A/m2{1 + coswt}

kislay

hey yahiyafirdous
you are correct but please explain equation (1)
how you got it
and i will give you salute

yahiyafirdous

Let x₂ is position of m₂ and that of m₁ is x₁

Let x is elongation in the spring.

Hence hence difference of position between them is elongation in the spring plus natural length.

Hence:

x₂-x₁=l₀+x

Please draw a figure, place both the particles on the same side of origin with their distances from origin x₁ and x_2.Now it will be clear how did it come about.If their difference is l₀, it means there is no elongation. If their difference is more than l₀, then there is elongation, if less than l₀ it means there is compression.

rajat

and lo and A are related by

lo = Aw[ sqroot{ (m1 + m2)/k } ]

am i ryt ?

iitkgp_bipin

x₁ = v₀t - A(1 - coswt)
dx₁/dt = v₀ - Awsinwt

At any instant position of centre of mass is v₀t .

(m₁x₁ + m₂x₂)/(m₁ + m₂) = v₀t

x₂ = v₀t + (m₁/m₂)A(1 - coswt)
dx₂/dt = v₀ + (m₁/m₂)Awsinwt

Now suppose that string is stretched by x.
Then x₂ - x₁ = original length + stretched length = l₀ + x

Differentiating both sides wrt t :

dx₂/dt - dx₁/dt = d(l₀ + x)/dt = dx/dt

(v₀ + (m₁/m₂)Awsinwt) - (v₀ - Awsinwt) = dx/dt

dx/dt = (m₁/m₂ + 1)Awsinwt

At extreme position total energy will be converted to PE.

(1/2)kl₀² = (1/2)k(m₁/m₂ + 1)²A²

l₀ = (m₁/m₂ + 1)A

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