Physics Laws of Motion
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New answer posted
a month agoContributor-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 agoNew answer posted
a month agoContributor-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
New answer posted
a month agoContributor-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 agoContributor-Level 9
For ball (a): I? = Δp? = 2mu
For ball (b): I? = Δp? = 2mu cos 45°
I? /I? = 2mu / (2mu cos 45°) = √2
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