We consider the “Cup + Collected Water” as our system.

• Mass of cup $\approx 0$ (inertia-less).
• Mass of water at time $t$: $m(t)$. At $t=0$, $m=0$.
• Rate of water addition: $\frac{dm}{dt} = r$.
• Velocity of the cup: $v_c$ (upwards).
• Velocity of water entering the cup: $v_{water}$. Since the problem states “negligible velocity relative to the cup,” we assume the water enters essentially at the cup’s velocity, so $v_{rel} \approx 0$.

The general equation of motion for a variable mass system is:

$$ F_{ext} + v_{rel} \frac{dm}{dt} = m \frac{dv}{dt} $$

Where:

• $F_{ext}$ is the net external force (Springs + Gravity).
• $v_{rel}$ is the relative velocity of the added mass with respect to the cup. The problem states water enters with “negligible velocity relative to the cup”, which implies $v_{rel} \approx 0$.
• Therefore, the thrust force term vanishes, and the equation simplifies to $F_{ext} = m a_c$.

Since the cup is inertia-less and at $t=0$ the mass of water $m=0$, the product $ma_c$ is zero. This implies the net external force must be zero at $t=0$.

$$ \Sigma F_{ext} = T_A – T_B – mg = 0 $$

At $t=0$, $m=0$, so:

$$ T_A – T_B = 0 \implies T_A = T_B $$

The equation $T_A – T_B – mg = 0$ holds continuously as the mass accumulates. To find the velocity, we differentiate this equilibrium condition with respect to time.

$$ \frac{d}{dt}(T_A – T_B – mg) = 0 $$ $$ \frac{dT_A}{dt} – \frac{dT_B}{dt} – g \frac{dm}{dt} = 0 $$

The rate of change of tension $\frac{dT}{dt} = k \frac{dx}{dt}$.

• Spring A (Top): The top end A moves up at $v_A$. The cup moves up at $v_c$. The rate of extension is $(v_A – v_c)$.
$$ \frac{dT_A}{dt} = k(v_A – v_c) $$
• Spring B (Bottom): The cup moves up at $v_c$. The bottom end B moves down at $v_B$. The rate of extension is $(v_c – (-v_B)) = v_c + v_B$.
$$ \frac{dT_B}{dt} = k(v_c + v_B) $$

Substitute the rates back into the differentiated force equation:

$$ k(v_A – v_c) – k(v_c + v_B) – gr = 0 $$ $$ k v_A – k v_c – k v_c – k v_B – gr = 0 $$ $$ k(v_A – v_B) – 2k v_c – gr = 0 $$ $$ 2k v_c = k(v_A – v_B) – gr $$ $$ v_c = \frac{k(v_A – v_B) – gr}{2k} $$

Given values:

• $k = 10 \text{ N/m}$
• $v_A = 10.0 \text{ cm/s} = 0.1 \text{ m/s}$
• $v_B = 8.0 \text{ cm/s} = 0.08 \text{ m/s}$
• $r = 1.0 \text{ g/s} = 0.001 \text{ kg/s}$
• $g = 10 \text{ m/s}^2$

Substitute into the expression:

$$ v_c = \frac{10(0.1 – 0.08) – (0.001)(10)}{2(10)} $$ $$ v_c = \frac{10(0.02) – 0.01}{20} $$ $$ v_c = \frac{0.2 – 0.01}{20} $$ $$ v_c = \frac{0.19}{20} = 0.0095 \text{ m/s} $$ $$ v_c = 0.95 \text{ cm/s} $$