Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 2

Answer Key - SA2 Physics Secondary 3 (Version 2)

Section A: MCQ

1. B (Acceleration due to gravity $g$ always acts downwards regardless of velocity).
2. A ($F = ma \rightarrow a = F/m$).
3. C ($W = F \times d = 15 \text{ N} \times 4 \text{ m} = 60 \text{ J}$).
4. C (Displacement has both magnitude and direction).
5. A ($W - f = ma \rightarrow (60 \times 10) - f = 60 \times 8 \rightarrow 600 - f = 480 \rightarrow f = 120 \text{ N}$).
6. B (Definition of equilibrium for moments).
7. C ($P = F_1/A_1 = F_2/A_2 \rightarrow 100/0.01 = F_2/0.1 \rightarrow F_2 = 1000 \text{ N}$).
8. C (Net force is zero; weight equals drag).
9. B ($P = F/A$; for a given weight, area determines pressure).
10. B ($v^2 = u^2 + 2as \rightarrow v^2 = 0 + 2gh \rightarrow v = \sqrt{2gh}$).

Section B: Structured

Question 11 (a) Diagram showing: Weight $W$ (down), Tension $T_1$ (up-left at $45^\circ$), Tension $T_2$ (up-right at $60^\circ$). [2] (b) $T_{1x} = T_1 \cos 45^\circ, T_{1y} = T_1 \sin 45^\circ$; $T_{2x} = T_2 \cos 60^\circ, T_{2y} = T_2 \sin 60^\circ$. [2] (c) $\sum F_x = 0 \rightarrow T_1 \cos 45^\circ = T_2 \cos 60^\circ \rightarrow T_2 = T_1 \frac{\cos 45^\circ}{\cos 60^\circ}$. $\sum F_y = W \rightarrow T_1 \sin 45^\circ + T_2 \sin 60^\circ = 0.5 \times 10 = 5 \text{ N}$. Substitute $T_2$: $T_1 \sin 45^\circ + (T_1 \frac{\cos 45^\circ}{\cos 60^\circ}) \sin 60^\circ = 5$. $T_1(0.707 + 0.707 \times 1.732) = 5 \rightarrow T_1(1.93) = 5 \rightarrow T_1 \approx 2.6 \text{ N}$. [2]

Question 12 (a) $W = F \times d = 120 \text{ N} \times 5 \text{ m} = 600 \text{ J}$. [2] (b) $\Delta GPE = mgh = 20 \times 10 \times 3 = 600 \text{ J}$. [2] (c) Energy loss = Work done - $\Delta GPE = 600 - 600 = 0 \text{ J}$. (Note: In this specific case, the force provided exactly matches the GPE gain, meaning friction is negligible or the scenario implies a frictionless plane). [3]

Question 13 (a) $10 \text{ m s}^{-2}$. [1] (b) Upon entering water, the ball experiences a large upward drag force and buoyancy. The net force decreases, causing the ball to decelerate. As speed decreases, drag decreases until drag + buoyancy = weight. [3] (c) Water is much denser than air. The drag force $F_d$ is proportional to fluid density. Therefore, the drag force reaches the value of the object's weight at a much lower velocity compared to air. [4]

Question 14 (a) Force $= 0.1 \text{ kg} \times 10 = 1 \text{ N}$. Distance from pivot $= 40 - 10 = 30 \text{ cm} = 0.3 \text{ m}$. Moment $= 1 \text{ N} \times 0.3 \text{ m} = 0.3 \text{ Nm}$. [2] (b) Clockwise moment $= 0.3 \text{ Nm}$. Weight of rule $= 0.15 \times 10 = 1.5 \text{ N}$. $1.5 \times d = 0.3 \rightarrow d = 0.2 \text{ m} = 20 \text{ cm}$ from pivot. Since the mass is at $10 \text{ cm}$ (left of pivot), the center of gravity must be to the right. Position $= 40 + 20 = 60 \text{ cm}$ mark. [3] (c) If pivot is at $50 \text{ cm}$, the center of gravity ($60 \text{ cm}$) is closer to the pivot. The rule becomes more stable if the base is widened, but as a balanced beam, it changes the equilibrium point. [2]

Question 15 (a) $a = F/m = 10/2 = 5 \text{ m s}^{-2}$. [2] (b) $F_{\text{net}} = 10 - 10 = 0 \text{ N}$. Acceleration $= 0$. The block moves at a constant velocity (the velocity it had at $t=3 \text{ s}$). [3] (c) $v$ at $3 \text{ s} = u + at = 0 + 5(3) = 15 \text{ m s}^{-1}$. $s_1$ (first $3 \text{ s}$) $= \frac{1}{2}at^2 = 0.5 \times 5 \times 3^2 = 22.5 \text{ m}$. $s_2$ (next $2 \text{ s}$) $= v \times t = 15 \times 2 = 30 \text{ m}$. Total $s = 22.5 + 30 = 52.5 \text{ m}$. [3]

Question 16 (a) $\Delta P = h\rho g = 0.15 \times 13600 \times 10 = 20,400 \text{ Pa}$. [3] (b) As gas is heated, pressure increases (since $V$ is constant). The gas pushes the mercury down on its side and up on the open side, increasing the height difference $h$. [4]

Question 17 (a) Energy cannot be created or destroyed, only transformed from one form to another. [1] (b) At top of loop, min speed $v_c = \sqrt{gr} = \sqrt{10 \times 0.5} = \sqrt{5} \text{ m s}^{-1}$. Total Energy at bottom $= \frac{1}{2}mv_0^2$. Total Energy at top $= \frac{1}{2}mv_c^2 + mg(2r) = \frac{1}{2}m(5) + m(10)(1) = 2.5m + 10m = 12.5m$. $\frac{1}{2}mv_0^2 = 12.5

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# Answer Key - SA2 Physics Secondary 3 (Version 2)

### Section A: MCQ
1. **B** (Acceleration due to gravity $g$ always acts downwards regardless of velocity).
2. **A** ($F = ma \rightarrow a = F/m$).
3. **C** ($W = F \times d = 15 \text{ N} \times 4 \text{ m} = 60 \text{ J}$).
4. **C** (Displacement has both magnitude and direction).
5. **A** ($W - f = ma \rightarrow (60 \times 10) - f = 60 \times 8 \rightarrow 600 - f = 480 \rightarrow f = 120 \text{ N}$).
6. **B** (Definition of equilibrium for moments).
7. **C** ($P = F_1/A_1 = F_2/A_2 \rightarrow 100/0.01 = F_2/0.1 \rightarrow F_2 = 1000 \text{ N}$).
8. **C** (Net force is zero; weight equals drag).
9. **B** ($P = F/A$; for a given weight, area determines pressure).
10. **B** ($v^2 = u^2 + 2as \rightarrow v^2 = 0 + 2gh \rightarrow v = \sqrt{2gh}$).

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### Section B: Structured

**Question 11**
(a) Diagram showing: Weight $W$ (down), Tension $T_1$ (up-left at $45^\circ$), Tension $T_2$ (up-right at $60^\circ$). [2]
(b) $T_{1x} = T_1 \cos 45^\circ, T_{1y} = T_1 \sin 45^\circ$; $T_{2x} = T_2 \cos 60^\circ, T_{2y} = T_2 \sin 60^\circ$. [2]
(c) $\sum F_x = 0 \rightarrow T_1 \cos 45^\circ = T_2 \cos 60^\circ \rightarrow T_2 = T_1 \frac{\cos 45^\circ}{\cos 60^\circ}$.
$\sum F_y = W \rightarrow T_1 \sin 45^\circ + T_2 \sin 60^\circ = 0.5 \times 10 = 5 \text{ N}$.
Substitute $T_2$: $T_1 \sin 45^\circ + (T_1 \frac{\cos 45^\circ}{\cos 60^\circ}) \sin 60^\circ = 5$.
$T_1(0.707 + 0.707 \times 1.732) = 5 \rightarrow T_1(1.93) = 5 \rightarrow T_1 \approx 2.6 \text{ N}$. [2]

**Question 12**
(a) $W = F \times d = 120 \text{ N} \times 5 \text{ m} = 600 \text{ J}$. [2]
(b) $\Delta GPE = mgh = 20 \times 10 \times 3 = 600 \text{ J}$. [2]
(c) Energy loss = Work done - $\Delta GPE = 600 - 600 = 0 \text{ J}$. (Note: In this specific case, the force provided exactly matches the GPE gain, meaning friction is negligible or the scenario implies a frictionless plane). [3]

**Question 13**
(a) $10 \text{ m s}^{-2}$. [1]
(b) Upon entering water, the ball experiences a large upward drag force and buoyancy. The net force decreases, causing the ball to decelerate. As speed decreases, drag decreases until drag + buoyancy = weight. [3]
(c) Water is much denser than air. The drag force $F_d$ is proportional to fluid density. Therefore, the drag force reaches the value of the object's weight at a much lower velocity compared to air. [4]

**Question 14**
(a) Force $= 0.1 \text{ kg} \times 10 = 1 \text{ N}$. Distance from pivot $= 40 - 10 = 30 \text{ cm} = 0.3 \text{ m}$.
Moment $= 1 \text{ N} \times 0.3 \text{ m} = 0.3 \text{ Nm}$. [2]
(b) Clockwise moment $= 0.3 \text{ Nm}$.
Weight of rule $= 0.15 \times 10 = 1.5 \text{ N}$.
$1.5 \times d = 0.3 \rightarrow d = 0.2 \text{ m} = 20 \text{ cm}$ from pivot.
Since the mass is at $10 \text{ cm}$ (left of pivot), the center of gravity must be to the right.
Position $= 40 + 20 = 60 \text{ cm}$ mark. [3]
(c) If pivot is at $50 \text{ cm}$, the center of gravity ($60 \text{ cm}$) is closer to the pivot. The rule becomes more stable if the base is widened, but as a balanced beam, it changes the equilibrium point. [2]

**Question 15**
(a) $a = F/m = 10/2 = 5 \text{ m s}^{-2}$. [2]
(b) $F_{\text{net}} = 10 - 10 = 0 \text{ N}$. Acceleration $= 0$. The block moves at a constant velocity (the velocity it had at $t=3 \text{ s}$). [3]
(c) $v$ at $3 \text{ s} = u + at = 0 + 5(3) = 15 \text{ m s}^{-1}$.
$s_1$ (first $3 \text{ s}$) $= \frac{1}{2}at^2 = 0.5 \times 5 \times 3^2 = 22.5 \text{ m}$.
$s_2$ (next $2 \text{ s}$) $= v \times t = 15 \times 2 = 30 \text{ m}$.
Total $s = 22.5 + 30 = 52.5 \text{ m}$. [3]

**Question 16**
(a) $\Delta P = h\rho g = 0.15 \times 13600 \times 10 = 20,400 \text{ Pa}$. [3]
(b) As gas is heated, pressure increases (since $V$ is constant). The gas pushes the mercury down on its side and up on the open side, increasing the height difference $h$. [4]

**Question 17**
(a) Energy cannot be created or destroyed, only transformed from one form to another. [1]
(b) At top of loop, min speed $v_c = \sqrt{gr} = \sqrt{10 \times 0.5} = \sqrt{5} \text{ m s}^{-1}$.
Total Energy at bottom $= \frac{1}{2}mv_0^2$.
Total Energy at top $= \frac{1}{2}mv_c^2 + mg(2r) = \frac{1}{2}m(5) + m(10)(1) = 2.5m + 10m = 12.5m$.
$\frac{1}{2}mv_0^2 = 12.5m \rightarrow v_0^2 = 25 \rightarrow v_0 = 5 \text{ m s}^{-1}$. [6]