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New answer posted
3 weeks agoContributor-Level 10
-mv cos 60? + 2mu = 0 => v = 4u
½m [v² + u² + 2uv cos 120? ] + ½mu² = mgx sin 60?
=> v² = (8/7)√3gx => ar = (4√3/7)g
∴ t = √ (2L * 7)/ (4√3g)
New answer posted
3 weeks agoContributor-Level 10
a = fr/m
∴ V = V? + at? = V? + (fr/m)t?
and α = 2Rfr/mR² = 2fr/mR
∴ ω = ω? - (2fr/mR)t?
∴ V = ωR ∴ v? + (fr/m)t? = ω? R - (2fr/m)t?
= (3fr/m)t? = V? = (fr/m)et?
∴ V = V? + V? /3 = 4V? /3
t? = mV? / (3μmg)
= V? /3μg
New answer posted
4 weeks agoContributor-Level 9
Initial velocity
Final velocity

Change in velocity
Momentum gain is along
Force experienced is along
Force experienced is in North-East direction.
New answer posted
a month agoContributor-Level 9
Apply conservation of momentum along y-axis, we can write
10v? - 10v? sin 30° = 0
⇒ v? = 20m/s
New answer posted
a month agoContributor-Level 10
Assuming the rope in the boy's hand is vertical and using a Free Body Diagram (FBD):
f? = T
R + T = 90 ⇒ R = 90 - T
For the piece of wood not to move, f? ≤ µR:
T ≤ 0.5 (90 - T) ⇒ T ≤ 30N
New answer posted
a month agoContributor-Level 10
The minimum force F? is calculated as:
F? = (μmg) / √* (1 + μ²)* = ( (1/√3) * 1 * 10 ) / √* (1 + (1/√3)²) = 5N
New answer posted
a month agoContributor-Level 10
For an elastic collision where C comes to rest, and the compression in the spring is maximum, the velocities of A and B are equal (v). Using the conservation of mechanical energy:
(1/2)mv? ² = 2 * (1/2)mv² + (1/2)kx²
This gives the maximum compression x as:
x = v? * √* (m / 2k)*
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