Physics

I have a doubt.

If a metal ball and plastic ball of same radius are $\text{gently dropped}$ from the top of a slope in such a way that both of them roll down after reaching the ground,which one will stop first?

#Mechanics

Easy Math Editor

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Comments

When the balls of same radius(and thus different mass) are dropped from the same height, they will have the same velocity when they strike the ground. Now, the velocity of rebounce for each will depend on the coefficient of restitution(e for short) between the colliding surface and the balls, thus each will have different velocity of rebounce.

Now, consider something else-when a ball is dropped from a height, it acquires some velocity just before hitting the surface. There will be some loss of energy due to the deformations and reformations occurring in the ball as it collides, and also as some energy is lost as heat or sound. If the collisions are not perfectly inelastic(that is complete loss of energy of ball does not take place), which indicates that $e \neq 0$ then the ball will rebound with some velocity less than or equal to it's hitting velocity. It will again collide and again rebound, with even lesser velocity(if e is not equal to 1)and in this way,it will keep on rebounding, with it's amplitude(i.e. the maximum height it reaches) becoming lesser and lesser.

Now,it will always have a velocity of hitting the surface(even after any number of collisions), and since $0<e \leq 1$ it will always have a velocity of rebounce. Thus, the ball will never really stop, and will continue to collide and rebound against the surface, with it's amplitude gradually becoming infinitesimally small, $amplitude \rightarrow 0$ but never becoming 0.

The above statements will be true even if we take air drag into account.

Your answer, will now depend on the following - If the value of coefficient of restitution between any ball and surface(which will depend on the nature of the two surfaces) is 0,then that ball will instantly come to rest, otherwise it will never come to rest.

Still, now if we move to the molecular level, when the ball collides with the floor, during it's deformation(or more specifically, during it's time of contact with the floor), formation of bonds between the two bodies(ball and floor) takes place, and these bonds provide resistance to the rebouncing of ball (just like frictional force) , and also breaking of these bonds leads to some loss of energy, but these forces can easily be ignored as they are too small as compared to the impulsive normal reactions which develop during it's collision.Nevertheless, ignoring or taking these forces into account won't change the answer to the question.

@Sai Ram I hope that your doubt is now resolved!

Abhijeet Verma - 5 years, 9 months ago

I have edited some parts of it.Can you please go through it and answer again ?

Sai Ram - 5 years, 9 months ago

@Sai Ram If pure rolling motion has been established at the ground(obviously due to friction) for both the balls, then none of them will ever stop.

(The above holds only if there is no air resistance)

Rolling motion is a motion in which no energy is lost due to any non conservative force.

Abhijeet Verma - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$