Physics Laws of Motion

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Statement-I: F? = mv²/r ≤ f? = µmg ⇒ v ≤ √ (µgR) = √ (0.2*9.8*2) = 1.98 m/s
v_cyclist = 7 km/h = 1.94 m/s. Since 1.94 m/s < 1.98 m/s, statement-I is correct.
Statement-II: v_max = √gR (tanθ+µ)/ (1-µtanθ) = .
v_min = √gR (tanθ-µ)/ (1+µtanθ) = √ (9.8*2 (tan45-0.2)/ (1+0.2tan45) = 3.65 m/s
v_cyclist = 18.5 km/h = 5.14 m/s. This is outside the safe range. Statement-II is incorrect.

New question posted

a month ago

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New answer posted

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

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V
Vishal Baghel

Contributor-Level 10

Let vel of A and B just after collision be VA & VB respectively.
m * 9 + 0 = m * VA + 2mVB
9 = VA + 2VB
again e = (VB - VA)/ (9-0) = 1
9 = VB - VA
From (i) & (ii)
VB = 6 m/s & VA = -3 m/s
Now, for B and C collision (completely inelastic):
2m * 6 + 0 = (2m + 2m)Vc
12m = 4mVc
Vc = 3 m/s

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a month ago

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A
alok kumar singh

Contributor-Level 10

a = F/m = (8î + 2? )
r = ut + ½ at²
r = 0 + ½ (8î + 2? ) 10²
r = 400î + 100?

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

For ball (a): I? = Δp? = 2mu
For ball (b): I? = Δp? = 2mu cos 45°
I? /I? = 2mu / (2mu cos 45°) = √2

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

If the block does not slide,

mg sin    θ μ m g c o s θ

  t a n θ μ d y d x μ x 2 0 . 5 x 1 m .          

Thus, y 1 2 4 = 0 . 2 5 m = 2 5 c m .

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

f = W = 0.5 * 10 = 5 N

N = F

For block not to slide,

  f μ N          

5 0 . 2 F F 2 5 N

New answer posted

a month ago

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P
Payal Gupta

Contributor-Level 10

mg – N = ma

N = m (g – a) = 60 * (10 – 1.8) = 60 * 8.2 = 492 N

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