Question

Two Solid spheres of same material having radii in the ratio $1:2$ are moving with same kinetic energy on a horizontal path. They are brought to rest by applying same retarding force. Then,
i) their initial momenta are in the ratio $1:2\sqrt{2}$
ii) distance travelled are in the ratio $1:1$ before coming to rest
iii) the times taken by them coming to rest is $1:2\sqrt{2}$

• A. i) and (ii) are correct
• B. ii) and (iii) are correct
• C. All the three are correct
• D. (i) and (iii) are correct

Answer 1

The correct option is C All the three are correct

1) m is $\propto r^3$

$\frac{m_I}{m_{II}}=\frac{1}{8}$

2)Kinetic energy $=\frac{1}{2}m{v}^2$

Given both spheres have same kinetic energy

$\frac{1}{2}{m}_I{v}_I^2$$=\frac{1}{2}{m}_{II}{v}_{II}^2$

$\Rightarrow \frac{v_I}{v_{II}}=2\sqrt{2}$

3) the momentum of spheres $p_I=mv,p_{II}=8m\times \frac{v}{2\sqrt{2}}$

$p_I:p_{II}=1:2\sqrt{2}$

4) $2as=v^2-{\mu }^2$

For I
$2\times \frac{F}{m}\times s=v^2$

$s_1=\frac{v^{2}m}{2F}$

For II

$s_2=\frac{v^2\left(8m\right)}{8\times 2F}$

$s_2=\frac{v^{2}m}{2F}$

$s_1:s_2=1:1$

5) $t_1=\frac{v_1}{a_1}=\frac{vm}{F}{t}_2=\frac{v}{2\sqrt{2}}\frac{8m}{F}$
$t_1:t_2=1:2\sqrt{2}$