Physics System of Particles and Rotational Motion
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New question posted
2 months agoNew answer posted
2 months agoContributor-Level 9
From conservation of momentum:
2*4 = 2v? + mv?
Given v? = 1 m/s (interpreted from intermediate steps)
8 = 2 (1) + mv?
mv? = 6 . (i)
From coefficient of restitution (e=1 for elastic collision):
e = (v? - v? )/ (u? - u? )
1 = (v? - v? )/ (4 - 0)
-1 = (v? - v? )/ (0 - 4) (as written in the image)
⇒ 4 = v? - 1
⇒ v? = 5 . (ii)
Put (2) in (1), m (5) = 6
m = 1.2kg
New answer posted
2 months agoContributor-Level 9
r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second
New answer posted
2 months agoContributor-Level 10
I? / I? = (MR²/2) / (mr²/4)
[ M = σπR², m = σπr² ]
= 2 (M/m) (R²/r²)
= 2 (σπR²/σπr²) (R²/r²)
= 2 (R²/r²)² = 2R? /r?
New answer posted
2 months agoContributor-Level 10
(a) I = ML²/12
(b) I = (2M)L²/3
(c) I = M (2L)²/12 = ML²/3
(d) I = (2M) (2L)²/3 = 8ML²/3
New answer posted
2 months agoContributor-Level 10
τ = Iα
I = ½MR² = ½ (10) (0.2)² = 0.2 kgm²
α = (ωf - ωi)/Δt = (0 - 600*2π/60)/10 = -2π rad/s²
τ = |Iα| = 0.2 * 2π = 0.4π = 4π * 10? ¹ Nm
New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
mg – T = ma. (1)
T * R = l . (2)
a = R. (3)
With the help of equations (1), (2) and (3), we get
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