2022 SQ

Physics 2022 SQ

SQ-1

(a) Fill in the blank: Archimedes’ principle states that a body partially or fully immersed in a fluid experiences an upthrust equal to the ……….. of the fluid displaced by the object.

(b) A submarine vessel of mass $M$ consists of two main parts: the inner pressure hull, surrounded by outer ballast tanks as shown in the simplified diagram. To control its buoyancy, the ballast tanks are filled with water from the sea. It is found that when the ballast tanks are completely emptied of seawater, $85 %$ of the volume $V$ of the submarine remains submerged underwater (Surfaced). For the submarine to completely submerge and be at rest, the ballast tanks needed to be filled with $370 m^{3}$ of seawater (Submerged). Density of seawater = $1025 {kg/m}^{3}$ . Volume of ballast $V_{b a l l a s t} = 520 m^{3}$
(i) Determine the volume $V$ of the submarine and the mass $M$ of the submarine.
(ii) Determine the acceleration of the submarine when the ballast tanks are fully filled. You can neglect the effects of water resistance.

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ANSWER

(a) Archimedes’ Principle

This first part is a fundamental concept. Let's get the definition just right.

Archimedes’ principle states that a body partially or fully immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced by the object.

Think of it like this: when you get into a bathtub, the water level rises. The buoyant force (or upthrust) pushing you up is exactly equal to the weight of that extra volume of water you've pushed out of the way. It's the fluid fighting back to reclaim its space!

(b) The Submarine Problem

Now for the calculations. This is where we apply the principle. The key to solving this is to analyze the forces acting on the submarine in each situation. When the submarine is at rest (either surfaced or submerged), the forces are balanced. The upward buoyant force must exactly equal the downward force of its total weight.

Let's define our variables first:

$M$ = Mass of the submarine structure
$V$ = Total volume of the submarine
$ρ_{s e a}$ = Density of seawater = $1025 {kg/m}^{3}$
$g$ = acceleration due to gravity (we'll let this cancel out in most steps, but we'll need it for the final acceleration calculation)
$F_{B}$ = Buoyant force (upthrust)
$W_{t o t a l}$ = Total weight of the submarine (structure + water in tanks)

The two main forces are:

Buoyant Force (upwards): $F_{B} = (Volume submerged) \times ρ_{s e a} \times g$
Total Weight (downwards): $W_{t o t a l} = (Total mass) \times g = (M + m_{w a t e r}) \times g$

(i) Determine the volume $V$ and mass $M$ of the submarine.

We have two scenarios given, which will give us two equations to solve for our two unknowns, $V$ and $M$ .

Scenario 1: Surfaced and at rest
The ballast tanks are empty, so the mass of water inside is zero ( $m_{w a t e r} = 0$ ).
$85 %$ of the volume is submerged, so the submerged volume is $0.85 V$ .
Since it's at rest, the forces are balanced:

F_{B} = W_{t o t a l} $ $ $ $ (Volume submerged) \times ρ_{s e a} \times g = M \times g

(0.85 V) \times ρ_{s e a} \times g = M \times g

We can cancel $g$ from both sides:

(0.85 V) \times ρ_{s e a} = M

Let's call this Equation (1).

Scenario 2: Fully submerged and at rest

The submarine is completely underwater, so the submerged volume is its total volume, $V$ .

The ballast tanks are filled with $370 m^{3}$ of seawater.

The mass of this water is $m_{w a t e r} = (Volume of water) \times ρ_{s e a} = 370 \times 1025 kg$ .

The total mass is now $M + m_{w a t e r}$ .

Again, it's at rest, so the forces are balanced:

F_{B} = W_{t o t a l}

(Total Volume) \times ρ_{s e a} \times g = (M + m_{w a t e r}) \times g

V \times ρ_{s e a} \times g = (M + 370 \times 1025) \times g

Cancelling $g$ again:

V \times ρ_{s e a} = M + (370 \times 1025)

Let's call this Equation (2).

Solving the Equations

Now we have a system of two equations. We can substitute Equation (1) into Equation (2).

V \times ρ_{s e a} = (0.85 V \times ρ_{s e a}) + (370 \times 1025)

Let's move the term with $V$ to the left side:

V \times ρ_{s e a} - 0.85 V \times ρ_{s e a} = 370 \times 1025

Factor out $V \times ρ_{s e a}$ :

(1 - 0.85) \times V \times ρ_{s e a} = 370 \times 1025

0.15 \times V \times 1025 = 370 \times 1025

We can divide both sides by 1025, which simplifies things nicely:

0.15 \times V = 370

V = \frac{370}{0.15} \approx 2466.7 m^{3}

So, the total volume of the submarine is approximately $V = 2470 m^{3}$ (to 3 significant figures).

Now we can find the mass $M$ using Equation (1):

M = 0.85 \times V \times ρ_{s e a}

M = 0.85 \times 2466.7 m^{3} \times 1025 {kg/m}^{3}

M \approx 2, 143, 833 kg

So, the mass of the submarine is approximately $M = 2.14 \times 10^{6} kg$ .

(ii) Determine the acceleration of the submarine when the ballast tanks are fully filled.

This is a different situation! The forces are no longer balanced, which means there will be a net force, causing acceleration according to Newton's Second Law ( $F_{n e t} = m a$ ).

First, let's figure out the forces in this new scenario.
The ballast tanks are fully filled. The total volume of the tanks is given as $V_{b a l l a s t} = 520 m^{3}$ .

Total mass ( $m_{t o t a l}$ ): This is the mass of the submarine structure ( $M$ ) plus the mass of the water in the full ballast tanks.
$m_{w a t e r_f u l l} = V_{b a l l a s t} \times ρ_{s e a} = 520 m^{3} \times 1025 {kg/m}^{3} = 533, 000 kg$ .

$m_{t o t a l} = M + m_{w a t e r_f u l l} = 2, 143, 833 kg + 533, 000 kg = 2, 676, 833 kg$ .
Total Weight (downwards, $W_{t o t a l}$ ):
$W_{t o t a l} = m_{t o t a l} \times g = 2, 676, 833 \times g$ .
Buoyant Force (upwards, $F_{B}$ ): Since the submarine will be accelerating downwards, it is fully submerged. The buoyant force depends only on the total volume $V$ , which doesn't change.
$F_{B} = V \times ρ_{s e a} \times g = 2466.7 m^{3} \times 1025 {kg/m}^{3} \times g = 2, 528, 333 \times g$ .

Now, we find the net force ( $F_{n e t}$ ). Let's define the downward direction as positive.

F_{n e t} = W_{t o t a l} - F_{B}

F_{n e t} = (2, 676, 833 \times g) - (2, 528, 333 \times g)

F_{n e t} = (2, 676, 833 - 2, 528, 333) \times g = 148, 500 \times g

Finally, we use Newton's Second Law, $F_{n e t} = m_{t o t a l} \times a$ .

148, 500 \times g = (2, 676, 833) \times a

Now we can solve for the acceleration, $a$ . The $g$ doesn't cancel this time! Let's use $g = 9.81 {m/s}^{2}$ .

a = \frac{148, 500 \times g}{2, 676, 833} = \frac{148, 500 \times 9.81}{2, 676, 833}

a \approx 0.544 {m/s}^{2}

The acceleration of the submarine is $0.54 {m/s}^{2}$ downwards.

SQ-3a

Two pieces of string labelled 1 and 2 are supporting a block of mass 4.0 𝑘𝑔. The angles are 𝜃_1 = 25° and 𝜃_2 = 40°. Calculate the tension 𝑇_1 and 𝑇_2 in the two strings 1 and 2 respectively.

ANSWER:

Step 1: Identify the Forces (Free-Body Diagram)

There are three forces pulling on this single point:

Tension $T_{1}$ : Pulling up and to the left along string 1.
Tension $T_{2}$ : Pulling up and to the right along string 2.
Weight (W): The force of gravity pulling the block straight down. This is the tension in the vertical string.

The weight of the block is calculated using Newton's second law, $F = m a$ . In this case, the acceleration is due to gravity, $g$ .

$W = m \times g$
$W = 4.0 kg \times 9.8 {m/s}^{2}$
$W = 39.2 N$

So, the downward pull is $39.2$ Newtons.

Step 2: Set Up a Coordinate System and Resolve Forces

Because the forces $T_{1}$ and $T_{2}$ are at an angle, we need to break them down into horizontal (x) and vertical (y) components. Let's define our standard coordinate system: +x is to the right, and +y is upwards.

For Tension $T_{1}$ : The angle $θ_{1} = 25^{\circ}$ is given with respect to the horizontal ceiling.
- Horizontal component ( $T_{1 x}$ ): $T_{1} \cos (25^{\circ})$ , pointing left (negative).
- Vertical component ( $T_{1 y}$ ): $T_{1} \sin (25^{\circ})$ , pointing up (positive).
For Tension $T_{2}$ : Be careful here! The angle $θ_{2} = 40^{\circ}$ is given with respect to the vertical wall. To use our standard sine and cosine functions relative to the horizontal axis, we need the angle with the horizontal. This angle is $90^{\circ} - 40^{\circ} = 50^{\circ}$ .
- Horizontal component ( $T_{2 x}$ ): $T_{2} \cos (50^{\circ})$ , pointing right (positive).
- Vertical component ( $T_{2 y}$ ): $T_{2} \sin (50^{\circ})$ , pointing up (positive).

Step 3: Apply the Equilibrium Conditions

Since the block is in equilibrium, the net force in both the x and y directions must be zero.

Condition 1: Sum of horizontal forces is zero ( $Σ F_{x} = 0$ )

The rightward forces must balance the leftward forces.

$T_{2 x} - T_{1 x} = 0$
$T_{2} \cos (50^{\circ}) - T_{1} \cos (25^{\circ}) = 0$

Let's call this Equation (1).

Condition 2: Sum of vertical forces is zero ( $Σ F_{y} = 0$ )

The upward forces must balance the downward force (the weight of the block).

$T_{1 y} + T_{2 y} - W = 0$
$T_{1} \sin (25^{\circ}) + T_{2} \sin (50^{\circ}) - 39.2 N = 0$

Let's call this Equation (2).

Step 4: Solve the System of Equations

Now we have two equations and two unknowns ( $T_{1}$ and $T_{2}$ ). We can solve this system. A good method is substitution.

From Equation (1), let's express $T_{1}$ in terms of $T_{2}$ :

$T_{1} \cos (25^{\circ}) = T_{2} \cos (50^{\circ})$
$T_{1} = T_{2} \frac{\cos (50^{\circ})}{\cos (25^{\circ})}$

Now, substitute this expression for $T_{1}$ into Equation (2):

$(T_{2} \frac{\cos (50^{\circ})}{\cos (25^{\circ})}) \sin (25^{\circ}) + T_{2} \sin (50^{\circ}) = 39.2$

Let's factor out $T_{2}$ :

$T_{2} (\frac{\cos (50^{\circ}) \sin (25^{\circ})}{\cos (25^{\circ})} + \sin (50^{\circ})) = 39.2$

Now, let's plug in the trigonometric values:

$\cos (50^{\circ}) \approx 0.6428$
$\cos (25^{\circ}) \approx 0.9063$
$\sin (25^{\circ}) \approx 0.4226$
$\sin (50^{\circ}) \approx 0.7660$

$T_{2} (\frac{0.6428 \times 0.4226}{0.9063} + 0.7660) = 39.2$
$T_{2} (0.2997 + 0.7660) = 39.2$
$T_{2} (1.0657) = 39.2$

$T_{2} = \frac{39.2}{1.0657} \approx 36.78 N$

Finally, we can find $T_{1}$ using the relationship we found earlier:

$T_{1} = T_{2} \frac{\cos (50^{\circ})}{\cos (25^{\circ})}$
$T_{1} = (36.78) \frac{0.6428}{0.9063}$
$T_{1} \approx 26.08 N$

Conclusion

The tension in string 1 is $T_{1} \approx 26 N$ .
The tension in string 2 is $T_{2} \approx 37 N$ .

SQ-3b

A block of mass 𝑚 slides from rest downwards along a rough slope of length 𝑑 long with kinetic coefficient of friction 𝜇𝑘 = 0.4. The angle of the slope is 𝜃.

(i) If 𝑚 = 1.2 kg, 𝑑 = 1.6 m and 𝜃 = 42∘, determine the magnitude of the velocity 𝑣 of the block at the bottom of the slope.

(ii) If 𝑚 = 1.2 kg, 𝑑 = 1.6 m and we require 𝑣 to be 4.0 m/s, determine 𝜃.

ANSWER:

The Big Idea: Work-Energy Theorem

The theorem states that the net work done on an object is equal to the change in its kinetic energy.

W_{n e t} = Δ K = K_{f i n a l} - K_{i n i t i a l}

The "net work" is the sum of the work done by all forces acting on the block as it slides down the slope.

The forces doing work are:

Gravity: It helps the block slide down.
Friction: It opposes the block's motion.

Since the block starts from rest, its initial kinetic energy, $K_{i n i t i a l}$ , is zero. The final kinetic energy is $K_{f i n a l} = \frac{1}{2} m v^{2}$ .

So our main equation is:

W_{g r a v i t y} + W_{f r i c t i o n} = \frac{1}{2} m v^{2}

Let's break down the work terms. As the block slides a distance $d$ down the slope:

Work by gravity is $W_{g r a v i t y} = (m g \sin θ) \times d$ . (The component of gravity along the slope is $m g \sin θ$ ).
Work by friction is $W_{f r i c t i o n} = - f_{k} \times d$ . (It's negative because it opposes motion). The force of kinetic friction is $f_{k} = μ_{k} N$ . On a slope, the normal force $N$ balances the perpendicular component of gravity, so $N = m g \cos θ$ . This gives us $f_{k} = μ_{k} m g \cos θ$ .

Putting it all together, our master equation is:

(m g \sin θ) d - (μ_{k} m g \cos θ) d = \frac{1}{2} m v^{2}

Part (i): Find the velocity $v$

Here, we are given:

$m = 1.2 kg$
$d = 1.6 m$
$θ = 42^{\circ}$
$μ_{k} = 0.4$

Let's use our master equation. An interesting thing happens if we simplify it first by dividing everything by $m$ :

g d (\sin θ - μ_{k} \cos θ) = \frac{1}{2} v^{2}

Notice the mass $m$ cancels out! The final speed of the block doesn't depend on its mass.

Now, we can solve for $v$ :

v = \sqrt{2 g d (\sin θ - μ_{k} \cos θ)}

Let's plug in the values ( $g = 9.8 {m/s}^{2}$ ):

$v = \sqrt{2 (9.8) (1.6) (\sin (42^{\circ}) - 0.4 \cos (42^{\circ}))}$
$v = \sqrt{31.36 (0.6691 - 0.4 \times 0.7431)}$
$v = \sqrt{31.36 (0.6691 - 0.2972)}$
$v = \sqrt{31.36 (0.3719)}$
$v = \sqrt{11.66}$
$v \approx 3.415 m/s$

So, the magnitude of the velocity at the bottom is approximately $3.4 m/s$ .

Part (ii): Find the angle $θ$

Now the situation is reversed. We know the desired final velocity and need to find the angle of the slope that achieves it.

We are given:

$m = 1.2 kg$
$d = 1.6 m$
$v = 4.0 m/s$
$μ_{k} = 0.4$

Let's go back to our simplified work-energy equation:

g d (\sin θ - μ_{k} \cos θ) = \frac{1}{2} v^{2}

Our goal is to solve for $θ$ .

Let's first isolate the part with the trig functions:

\sin θ - μ_{k} \cos θ = \frac{v^{2}}{2 g d}

Now, let's calculate the value of the right-hand side:

\frac{(4.0)^{2}}{2 (9.8) (1.6)} = \frac{16}{31.36} \approx 0.5102

So we need to solve the trigonometric equation:

\sin θ - 0.4 \cos θ = 0.5102

This is a bit tricky to solve directly. A standard technique is to convert the left side into a single sine function of the form $R \sin (θ - α)$ .

Where $R = \sqrt{1^{2} + (- 0.4)^{2}} = \sqrt{1.16} \approx 1.077$ and $α = \arctan (\frac{0.4}{1}) \approx {21.8}^{\circ}$ .

So the equation becomes:

1.077 \sin (θ - {21.8}^{\circ}) = 0.5102

\sin (θ - {21.8}^{\circ}) = \frac{0.5102}{1.077} \approx 0.4737

Now we can take the inverse sine (arcsin):

θ - {21.8}^{\circ} = \arcsin (0.4737)

θ - {21.8}^{\circ} \approx {28.27}^{\circ}

Finally, solving for $θ$ :

θ = {28.27}^{\circ} + {21.8}^{\circ} = {50.07}^{\circ}

So, the required angle is approximately $50^{\circ}$ .