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As physics p1 MCQS YEARLY ONLY.

Thought blocker

Thought blocker

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Hadi Murtaza said:
Q.8
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(2m)(u) + (m)(-u) = (2m)(v₁) + (m)(v₂)
2mu - mu = 2mv₁ + mv₂
mu = 2mv₁ + mv₂
u = 2v₁ + v₂ . . . . . (i)

e = 1 , as collision is elastic
e = (v₂ - v₁)/(u₁ - u₂)
1 = (v₂ - v₁)/(u₁ - u₂)
v₂ - v₁ = u₁ - u₂
v₂ - v₁ = u - (-u)
v₂ - v₁ = 2u
v₂ = 2u + v₁

Substitute in (i)
u = 2v₁ + (2u + v₁)
u = 3v₁ + 2u
3v₁ = -u
[ v₁ = -u/3 ]

v₂ = 2u + v₁
v₂ = 2u - u/3
[ v₂ = 5u/3 ]

So answer: A
Click to expand...
I think this is a long method...
 
Thought blocker

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forever_chocoholic said:
but momentum is conserved in elastic collision rite?
Click to expand...
Well see this :¬
Pi = pf right ?
so Pi = 2mu - mu = mu
and for option A
-2mu/3 + 5mu/3 = mu so momentum is conserved. Pi = Pf
for B -2mu/6 + 2mu/3 = 4mu/6 so momentum is not conserved Pi ≠ Pf
for C 2mu/6 + 2mu/3 = mu so momentum is conserved Pi =Pf
for D (2m + m) * u/3 = mu so momentum is conserved Pi =Pf
We are asked for elastic collision hence C and D are inelastic as C is moving in 1 direction i.e is +ve and D is stick.. so answer is A :)
 
Last edited:
Mohammed Siddiq

Mohammed Siddiq

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Thought blocker said:
hmm.. let's see.. it is given : 500 lines in 1 mm, therefore, 1 line = 0.001/500 = 2 * 10^-6
the general equation to find the maximum no. of order, i.e the maximum no. of images = d/lamba = (2 * 10^-6)/(600*10^-9) = 3.33 so, the maximum no. or order is 3, but, you have to consider the other 3 images at position Z, hmm.. firstly, zero order is at position Y, so, that's one image, then there are three orders, i.e, three images at position X as well as at position Z, therefore, altogether there are 7 images, 3 images at position X, 1 image at position Y, and 3 images at position Z.. I hope you understand my explanations.. heheh.. :)
Click to expand...
Ah, yes. I forgot about the bright fringe that forms at Y. Thanks amico.
 
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