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New answer posted

a month ago

0 Follower 2 Views

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

New answer posted

a month ago

0 Follower 2 Views

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

0 Follower 7 Views

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

0 Follower 5 Views

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

0 Follower 3 Views

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

New answer posted

a month ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

N = mg + F sin 60°

flim=Fcos60°μN=Fcos60°

μmg=F [cos60°μsin60°]

F=μmgcos60°μsin60°=133*3*1012133*32=10/31216=10/31/3=10N=3* (103)N

New answer posted

a month ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

ax=Fxm=202=10m/s2

Sx=12axt2=12*10*102

New answer posted

a month ago

0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.

New answer posted

a month ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

P=2Km

n2=m1m2=12

n = 1

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