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New 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
New answer posted
a month agoContributor-Level 10
Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.
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