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A small, light ball and a larger, heavier ball are released from the top of a slope.

Which will move further? which will come down faster?

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Like this it looks too much like a homework. What does confuse you in this problem? Why you can not solve it yourself? –  Maksim Zholudev Jan 15 '12 at 18:52

1 Answer 1

Assumptions: The slope is not slippery, the balls have similar mass distribution, negligible air drag.

The angle of the slope is $\theta$, and mass of one of the ball is $m$, and its radius is $R$.

$a=\alpha R$ where $\alpha$ is the angular acceleration of the ball, and torque by static friction $f$ is $\tau = f R$, and $ mg \sin \theta - f = ma$. we also know $\tau = \beta mR^2 \alpha$ where $\beta$ is some constant (depends only on the distribution of mass).

Solving for $a$ yields $a=\frac{g \sin\theta}{1+\beta}$, independent of $m$ and $R$. Therefore, both balls should have same acceleration (and therefore same velocity and displacement).

However, if there's air drag, their acceleration depends on the radius and the mass of the ball, so not enough information is given for this case. If the balls are of the same material (same density $\rho$), the larger one comes down faster than the smaller one. Here's a proof...

After very long time, since air drag is proportional to the velocity $f_d=6\pi\eta v R$ ($\eta$ = viscosity of air), the balls will eventually reach their terminal velocities (denoted as $v_t$) and their accelerations are close enough to zero. With setting $f_d = 6\pi\eta v_t R = m g \sin\theta = \frac43 \pi \rho R^3$, we get the fact that the terminal velocity is proportional to the square of the radius. Therefore the larger one comes down faster.

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