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3 - (0625-S 2017-Paper 2 (Extended)/1-Q3) - FORCES AND MOTION A skydiver reache...
Jul 19, 2024
Solution
a
When the skydiver opens the parachute, the increased air resistance causes a significant deceleration. This deceleration is an upward acceleration because it opposes the downward motion
b
The correct answer is *B*
Question 2: Acceleration and Distance of Toy Car
a
To find the acceleration during the first two seconds, we use the slope of the speed-time graph. The slope is given by the change in speed divided by the change in time: a=ΔvΔt=3.8 m/s2 s=1.9 m/s2a = \frac{\Delta v}{\Delta t} = \frac{3.8 \, \text{m/s}}{2 \, \text{s}} = 1.9 \, \text{m/s}^2
b
To find the total distance traveled, we calculate the area under the speed-time graph. The graph forms a triangle during the first two seconds: Area=12×base×height=12×2 s×3.8 m/s=3.8 m\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \, \text{s} \times 3.8 \, \text{m/s} = 3.8 \, \text{m}
c
The correct answer is not listed in the table provided
Question 3: Resultant Force on Boat
a
When a boat is traveling at a steady speed in a straight line, the forces acting on it are balanced. This means the resultant force is zero
b
The correct answer is *D*
Answer
B, Not listed, D
Key Concept
Newton's First Law: An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
Explanation
For the skydiver, the opening of the parachute causes an upward acceleration due to increased air resistance. For the toy car, the acceleration and distance calculations are based on the speed-time graph. For the boat, a steady speed implies balanced forces, resulting in zero net force.
Solution
a
Identify the pivot point and the forces acting on the wheelbarrow. The wheel acts as the pivot. The forces are the weight of the sack and barrow (200 N) acting downward at a distance of 15 cm from the pivot, and the vertical force exerted by the man acting upward at a distance of 45 cm from the pivot
b
Apply the principle of moments (torque) about the pivot point. For the system to be in equilibrium, the sum of the clockwise moments must equal the sum of the counterclockwise moments. The moment due to the weight of the sack and barrow is 200 N×0.15 m200 \, \text{N} \times 0.15 \, \text{m}. The moment due to the force exerted by the man is F×0.45 mF \times 0.45 \, \text{m}
c
Set up the equation for equilibrium: 200 N×0.15 m=F×0.45 m200 \, \text{N} \times 0.15 \, \text{m} = F \times 0.45 \, \text{m}
d
Solve for FF: F=200 N×0.15 m0.45 m=66.67 NF = \frac{200 \, \text{N} \times 0.15 \, \text{m}}{0.45 \, \text{m}} = 66.67 \, \text{N}
Answer
C
Key Concept
Static Equilibrium: For an object to be in static equilibrium, the sum of the forces and the sum of the moments (torques) acting on it must be zero. Equation: ∑F=0\sum F = 0 and ∑τ=0\sum \tau = 0
Explanation
By applying the principle of moments about the pivot point, we can determine the vertical force exerted by the man.
Question 2: Lifting Force Required to Lift the Bridge
a
Identify the pivot point and the forces acting on the bridge. The bridge is pivoted at one end, and the weight of the bridge (10,000 N) acts downward at the center of the bridge (2.0 m from the pivot). The lifting force is applied at the other end (4.0 m from the pivot)
b
Apply the principle of moments (torque) about the pivot point. For the bridge to start lifting, the moment due to the lifting force must equal the moment due to the weight of the bridge. The moment due to the weight of the bridge is 10,000 N×2.0 m10,000 \, \text{N} \times 2.0 \, \text{m}. The moment due to the lifting force is F×4.0 mF \times 4.0 \, \text{m}
c
Set up the equation for equilibrium: 10,000 N×2.0 m=F×4.0 m10,000 \, \text{N} \times 2.0 \, \text{m} = F \times 4.0 \, \text{m}
d
Solve for FF: F=10,000 N×2.0 m4.0 m=5,000 NF = \frac{10,000 \, \text{N} \times 2.0 \, \text{m}}{4.0 \, \text{m}} = 5,000 \, \text{N}
Answer
B
Key Concept
Static Equilibrium: For an object to be in static equilibrium, the sum of the forces and the sum of the moments (torques) acting on it must be zero. Equation: ∑F=0\sum F = 0 and ∑τ=0\sum \tau = 0
Explanation
By applying the principle of moments about the pivot point, we can determine the lifting force required to lift the bridge.
Question 3: Change in Depth of the Submarine
a
Identify the given information: The density of water is 1.0×103 kg/m31.0 \times 10^3 \, \text{kg/m}^3, and the change in pressure is 0.10 MPa=0.10×106 Pa0.10 \, \text{MPa} = 0.10 \times 10^6 \, \text{Pa}
b
Use the hydrostatic pressure formula: ΔP=ρgΔh\Delta P = \rho g \Delta h where ΔP\Delta P is the change in pressure, ρ\rho is the density of water, gg is the acceleration due to gravity (9.8 m/s29.8 \, \text{m/s}^2), and Δh\Delta h is the change in depth
c
Rearrange the formula to solve for Δh\Delta h: Δh=ΔPρg\Delta h = \frac{\Delta P}{\rho g}
d
Substitute the given values: Δh=0.10×106 Pa1.0×103 kg/m3×9.8 m/s2=10.2 m\Delta h = \frac{0.10 \times 10^6 \, \text{Pa}}{1.0 \times 10^3 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2} = 10.2 \, \text{m}
Answer
B
Key Concept
Hydrostatic Pressure: The pressure in a fluid at a given depth is given by the hydrostatic pressure formula ΔP=ρgΔh\Delta P = \rho g \Delta h, where ΔP\Delta P is the change in pressure, ρ\rho is the fluid density, gg is the acceleration due to gravity, and Δh\Delta h is the change in depth.
Explanation
By using the hydrostatic pressure formula and substituting the given values, we can determine the change in depth of the submarine.
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