The mass of the object can be found using the conservation of energy:

$$
frac{1}{2}mv_1^2 + PE_1 = frac{1}{2}mv_2^2 + PE_2
$$

Setting $v_1 = 0$ (the particle starts at rest) and rearranging the equation, we obtain

$$
frac{1}{2}mv_2^2 = PE_1 – PE_2
$$

Realizing that $Delta PE = PE_2 – PE_1 = qDelta V$, or we say $PE_1 – PE_2 = -qDelta V$, we can rewrite the equation as

$$
frac{1}{2}mv_2^2 = -qDelta V
$$

We know $v_2 = 21$ m/s, $q = 9.2times10^{-5}$ C, and $Delta V = -45$ V, so solving for $m$ and substituting the know values yield

$$
begin{align*}
m &= -frac{2qDelta V}{v^2}\
&= -frac{2(9.2times10^{-5};mathrm{C})(-45;mathrm{V})}{(21;mathrm{m/s})^2}\
&= 1.9times10^{-5};mathrm{kg}
end{align*}
$$

where we used the units conversion (1 C)(1 V) = 1 J = 1 kg$cdot;mathrm{m^2}/mathrm{s}^2$.

$$
1.9times10^{-5};mathrm{kg}
$$

The potential difference can be found using the conservation of energy:

$$
frac{1}{2}m_ev_1^2 + PE_1 = frac{1}{2}m_ev_2^2 + PE_2
$$

We set $v_1 = 0$ (as the particle starts from rest) and rearrange the equation:

$$
PE_2 – PE_1 = -frac{1}{2}m_ev_2^2
$$

The next step is to express the change in the electric potential energy as $Delta PE = PE_2 – PE_1 = (-e)V$, where $-e$ is the electron’s charge and $V$ is the required potential difference. So we have

$$
-eV = -frac{1}{2}m_ev_2^2
$$

Solving for $V$ and substituting the known values ($-e$ and $m_e$ are the known charge and mass of the electron respectively), we find

$$
begin{align*}
V &= frac{m_ev_2^2}{2e}\
&= frac{(9.1times10^{-31};mathrm{kg})(4.8times10^{4};mathrm{m/s})^2}{2(1.6times10^{-19};mathrm{C})}\
&= 6.6;mathrm{mV}
end{align*}
$$

(remember that $1;mathrm{kg}cdotmathrm{(m/s)}^2$ = 1 J = (1 C)(1 V).)

The object’s final speed can be found using energy conservation. To do so, we first solve the energy conservation equation for the final kinetic energy:

$$
begin{align*}
frac{1}{2}mv_2^2 &= frac{1}{2}mv_1^2 + PE_1 – PE_2\
&= frac{1}{2}mv_1^2 + q(V_1 – V_2)
end{align*}
$$

Then we solve for $v_2$ and substitute the known values:

$$
begin{align*}
frac{1}{2}mv_2^2 &= frac{1}{2}mv_1^2 + q(V_1 – V_2)\
v_2^2 &= v_1^2 + frac{2q(V_1 – V_2)}{m}\
v_2 &= sqrt{v_1^2 + frac{2q(V_1 – V_2)}{m}}\
&= sqrt{(2.6;mathrm{m/s})^2 + frac{2(8.0times10^{-19};mathrm{C})(7500;mathrm{V})}{4.2times10^{-15};mathrm{kg}}}\
&= 3.1;mathrm{m/s}
end{align*}
$$