AP Physics 1- 8.4 Fluids and Conservation Laws- Exam Style questions - FRQs- New Syllabus

(a)(i)
Air particles collide with the top of each block and exert downward forces on the block.
Water particles collide with the bottom and sides of each block, and the collisions from below exert upward forces on the block.

Because pressure in the water increases with depth, the water particles below the block exert a greater upward force than the air particles above exert downward force.
Therefore, the net force from the surrounding fluid particles is upward, which is the buoyant force.

Another way to say this is that deeper fluid pushes more strongly, so the bottom of the block is pushed harder than the top.

(a)(ii)
Block \( A \) has the greater density.

Since the blocks have identical dimensions, the one with greater density must have greater mass and weight.
For floating objects, the buoyant force equals the weight, so the heavier block must displace more water.

That means Block \( A \) sits deeper in the water than Block \( B \). As the water level drops, the lower-floating block reaches the bottom first. Since \( A \) touches the bottom while \( B \) is still above it, \( A \) must be denser.

(b)(i)
Apply Bernoulli’s equation between the top surface of the water and the pipe exit.

At the top surface: pressure \( = P_{\text{atm}} \), speed \( = v_s \), height \( = h \).

At the pipe exit: pressure \( = P_{\text{atm}} \), speed \( = v_p \), height \( = 0 \).

So,

\( P_{\text{atm}} + \rho gh + \dfrac{1}{2}\rho v_s^2 = P_{\text{atm}} + \dfrac{1}{2}\rho v_p^2 \)

Cancel \( P_{\text{atm}} \) and divide by \( \rho \):

\( gh + \dfrac{1}{2}v_s^2 = \dfrac{1}{2}v_p^2 \)

Therefore,

\( v_p^2 = v_s^2 + 2gh \)

\( \boxed{v_p=\sqrt{v_s^2+2gh}} \)

(b)(ii)
Use the continuity equation. The rate at which volume leaves the top surface must equal the rate at which volume exits the pipe.

Area of top surface \( = \pi R^2 \)
Area of pipe \( = \pi r^2 \)

Thus,

\( A_{\text{top}}v_s = A_{\text{pipe}}v_p \)

\( \pi R^2 v_s = \pi r^2 v_p \)

Solving for \( v_p \),

\( \boxed{v_p=\dfrac{R^2}{r^2}v_s} \)

This makes sense because if the top area is much larger than the pipe area, the surface moves slowly while the exiting water moves much faster.

(b)(iii)
If \( R \gg r \), then from part \( \mathrm{(b)(ii)} \),

\( v_s = \dfrac{r^2}{R^2}v_p \)

Because \( \dfrac{r^2}{R^2} \) is very small, \( v_s \) is much smaller than \( v_p \), so \( v_s^2 \) is negligible compared with \( v_p^2 \).

Then in part \( \mathrm{(b)(i)} \),

\( v_p=\sqrt{v_s^2+2gh}\approx\sqrt{2gh} \)

So the claim is justified:

\( \boxed{v_p \approx \sqrt{2gh}} \)

(c)
The speed \( v_s \) of the surface decreases over time.

From part \( \mathrm{(b)(i)} \), as water drains out, the height \( h \) decreases, so \( v_p \) decreases.

From part \( \mathrm{(b)(ii)} \),

\( v_s = \dfrac{r^2}{R^2}v_p \)

In Tank \( Z \), the tank gets narrower toward the top, so as the water level drops, the radius \( R \) of the surface increases. That makes the factor \( \dfrac{r^2}{R^2} \) decrease.

Therefore, both effects work in the same direction:

• \( h \downarrow \Rightarrow v_p \downarrow \)
• \( R \uparrow \Rightarrow v_s = \dfrac{r^2}{R^2}v_p \downarrow \)

So the downward speed of the surface gets smaller with time.

\( \boxed{v_s \text{ decreases over time}} \)