$$
begin{align*}
triangle t=frac{triangle t_0}{sqrt{1-frac{v^2}{c^2}}}
end{align*}
$$

where $triangle t$ is the time measured in the lab frame of reference, $triangle t_0$ is the proper time, $v$ is the speed of the moving frame of reference and $c$ is the speed of light in vacuum. We see from the equation above that the time measured in the laboratory frame of reference $triangle t$ is bigger than the proper time $triangle t_0$ (this is why the effect is called time dilation). We conclude that the right answer is: B). The moving clock must run faster, ie. it will measure a greater time interval than that of a clock that is stationary.

$$
t=frac{1text{ s}}{sqrt{1-0.8^2}} =1.67text{ s}.
$$

$$
5text{ cm} = 10text{ cm}sqrt{1-frac{v^2}{c^2}}
$$

which gives

$$
v=csqrt{frac{3}{4}} = 0.87c.
$$

$$
E_1 = frac{mc^2}{sqrt{1-frac{v_1^2}{c^2}}} = 2.25times 10^{17} text{ J}.
$$

$$
E_2 = frac{mc^2}{sqrt{1-frac{v_2^2}{c^2}}} = 2.6times 10^{17}text{ J}.
$$

b) Using classical expression we get

$$
KE_1 = frac{1}{2}mv_1^2 =12500text{ J}
$$

$$
KE_2 = frac{1}{2}mv_2^2 = 2.81times 10^{16}text{ J}.
$$

c) Adding rest energy to classically calculated kinetic energy gives

$$
E_1’=2.25times10^{17}text{ J}
$$

$$
E_2′ = 2.53times10^{17}text{ J}.
$$

We see perfect matching for $v_1$ of the total energies and not so well matching for $v_2$. This is significant because it tells us that classical and relativistic expressions for speeds much smaller than $c$ give exactly the same results.