Trial Q (1 Viewer)

[krypticlemonjuice]

Can someone show their working for this q please??
The answer is A

 

[Unovan]

i thought bro was taking a piss at first glance and you had to calculate the trajectory of that shit

 

[wizzkids]

The mass loses kinetic energy and gains GPE as it moves from A to B. Let the velocity at the bottom be v₁ and the velocity at the top be v₂
The change of gpe is U = 2mgr.
The change of KE is U = -1/2 m (v₁² - v₂²)
Equate the two expressions and eliminate the common factor "m" you get:
2gr = -1/2 (v₁² - v₂²)
With a little bit of factorisation you can show that:
v₁ - v₂ = -4gr / (v₁ + v₂)

 

[krypticlemonjuice]

The mass loses kinetic energy and gains GPE as it moves from A to B. Let the velocity at the bottom be v₁ and the velocity at the top be v₂
The change of gpe is U = 2mgr.
The change of KE is U = -1/2 m (v₁^2 - v₂^2)
Equate the two expressions and eliminate the common factor "m" you get:
2gr = -1/2 (v₁^2 - v₂^2)
With a little bit of factorisation you can show:
v₁ - v₂ = -4gr / (v₁ + v₂)

Ohhh thanks so much. that makes sense