Multiple Choice Kinematics Problem

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

$α = v\frac{dv}{ds}$ and K is the kinetic energy .


In all the four cases $α = v\frac{dv}{ds} = mgsinθ$ where θ is the angle which the normal to the body makes with the vertical .

Fig 1) K increases and tangential acceleration α decreases ,so matches with P)

Fig 2) K decreases and tangential acceleration α also decreases ,so matches with S)

Fig 3) K increases and tangential acceleration α also increases ,so matches with R)

Fig 4) K decreases and tangential acceleration α increases ,so matches with Q)

According to me the correct option is (C)

Is this the correct option ?

Many thanks

 

[collinsmark]

Homework Statement

Homework Equations

The Attempt at a Solution

$α = v\frac{dv}{ds}$ and K is the kinetic energy .


In all the four cases $α = v\frac{dv}{ds} = mgsinθ$ where θ is the angle which the normal to the body makes with the vertical .

I think you accidentally added an m in there that doesn't belong.

Fig 1) K increases and tangential acceleration α decreases ,so matches with P)

Fig 2) K decreases and tangential acceleration α also decreases ,so matches with S)

Fig 3) K increases and tangential acceleration α also increases ,so matches with R)

Fig 4) K decreases and tangential acceleration α increases ,so matches with Q)

According to me the correct option is (C)

Is this the correct option ?

Yes, your choice looks correct to me.

In case you missed it though, there is another way to express the \alpha = v \frac{dv}{ds}. Can you represent v in terms of ds and dt? if so, replace v with that in your equation and see what happens. It should then be quite clear why the g \sin \theta makes perfect sense.

 

[Vibhor]

Yes, your choice looks correct to me.

Thanks :)

In case you missed it though, there is another way to express the \alpha = v \frac{dv}{ds}. Can you represent v in terms of ds and dt? if so, replace v with that in your equation and see what happens. It should then be quite clear why the g \sin \theta makes perfect sense.

v = \frac{ds} {dt} and α = \frac{dv}{dt} which is the tangential acceleration .

Is this what you are suggesting ?

 

[collinsmark]

v = \frac{ds} {dt} and α = \frac{dv}{dt} which is the tangential acceleration .

Is this what you are suggesting ?

That's it. It was just to point out that \alpha is the ball's acceleration.

 

[collinsmark]

And yes, just to avoid ambiguity, \alpha is the tangential component of acceleration, as you have correctly suggested. (There's also a normal component, but that's not relevant for this problem.)