Question

If a sphere with twice the radius and four times the mass had been used; what would have been its final speed?

• A. $\frac{5}{7}{v}_0$
• B. $\frac{2}{7}{v}_0$
• C. $\frac{7}{5}{v}_0$
• D. $\frac{7}{2}{v}_0$

Answer 1

The correct option is B $\frac{2}{7}{v}_0$

In the truck frame, the initial velocity of the sphere is $v_o$ in the backward direction.

Let $v$ and $w$ be the linear and angular velocity of the sphere when pure rolling starts.

$v=R^{\prime }w$

${\tau }_o=I\alpha$

$f{R}^{\prime }=\frac{2}{5}{M}^{\prime }{R}^{\prime 2}\alpha ⟹\alpha R^{\prime }=\frac{5f}{2M^{\prime }{R}^{\prime }}$

$w=0+\alpha t$

$w=\frac{5f}{2M^{\prime }{R}^{\prime }}t$ ..............(1)

Also, $a=\frac{f}{M^{\prime }}$

$v=v_o-at⟹v=v_o-\frac{ft}{M^{\prime }}$

Using (1), $⟹v=v_o-\frac{2R^{\prime }w}{5}$

$v=v_o-\frac{2v}{5}$ ......................(2)

Equation (2) is independent of mass and radius.

Thus $v=\frac{5}{7}{v}_o$ w.r.t truck

Now in the ground frame, final velocity of sphere $v^{\prime }=v_o-v=\frac{2}{7}{v}_o$