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.)