Physics 10 F16 Homework 2 Name Identify Known And Unknown
Phy 10 F16 Hw 2nameidentify Known And Unknown Wr
Identify known and unknown, write down relevant equations, and solve the problems. Do not forget units! Write each answer in an orderly manner!
Paper For Above instruction
Introduction
This paper aims to address a series of physics problems involving kinematics, momentum, energy, forces, and simple machines, with a focus on identifying known and unknown quantities, applying relevant equations, and providing detailed step-by-step solutions. The problems cover real-world scenarios such as airplane motion, billiard balls, projectile motion, work and power calculations, levers, and pulleys, offering a comprehensive review of fundamental physics principles.
Problem 1: Final Speed of an Airplane with Wind Influence
Known:
- Airplane speed relative to ground: \(v_{a} = 150\, \text{km/h}\), aligned north-south.
- Wind speed: \(v_{w} = 20\, \text{km/h}\), in east-west direction.
- Directions: Wind is perpendicular to the plane's motion.
Unknown:
- Final speed of airplane considering wind effects, \(v_{f}\).
Equation:
- Use vector addition: \(\vec{v}_{f} = \sqrt{v_{a}^{2} + v_{w}^{2}}\).
Calculation:
- Convert speeds to m/s:
\(150\, \text{km/h} = \frac{150 \times 1000}{3600} \approx 41.67\, \text{m/s}\),
\(20\, \text{km/h} = \frac{20 \times 1000}{3600} \approx 5.56\, \text{m/s}\).
- Final speed:
\[
v_{f} = \sqrt{(41.67)^2 + (5.56)^2} \approx \sqrt{1736.11 + 30.86} \approx \sqrt{1766.97} \approx 42.05\, \text{m/s}.
\]
- Converting back to km/h:
\[
v_{f} \approx 42.05 \times \frac{3600}{1000} \approx 151.4\, \text{km/h}.
\]
Answer:
The airplane’s final speed is approximately 151.4 km/h considering the wind effect.
Problem 2: Momentum of a billiard ball and comparison with a bullet
Known:
- Mass of billiard ball: \(m_{b} = 160\, \text{g} = 0.16\, \text{kg}\).
- Velocity of billiard ball: \(v_{b} = 12\, \text{m/s}\).
- Mass of bullet: \(m_{bullet} = 7.5\, \text{g} = 0.0075\, \text{kg}\).
Unknown:
- Momentum of billiard ball: \(p_b\).
- Velocity of bullet \(v_{bullet}\) where its momentum equals that of billiard ball.
Equation:
- Momentum: \(p = m v\).
Calculations:
- Momentum of billiard ball:
\[
p_b = m_b v_b = 0.16\, \text{kg} \times 12\, \text{m/s} = 1.92\, \text{kg·m/s}.
\]
- To find \(v_{bullet}\):
\[
p_{bullet} = p_b \Rightarrow m_{bullet} v_{bullet} = 1.92.
\]
- Solving for \(v_{bullet}\):
\[
v_{bullet} = \frac{1.92}{0.0075} \approx 256\, \text{m/s}.
\]
Answer:
The momentum of the billiard ball is 1.92 kg·m/s, and the bullet must travel at approximately 256 m/s to have the same momentum.
Problem 3: Tossing the billiard ball upward and analyzing motion
Known:
- Initial velocity: \(v_0 = 40\, \text{m/s}\).
- Acceleration due to gravity: \(g = 9.8\, \text{m/s}^2\).
- Time intervals: \(t = 2\, \text{s}\), \(6\, \text{s}\).
Unknown:
a) Momentum after 2 s: \(p_{2}\).
b) Momentum at highest point: \(p_{max}\).
c) Momentum when returning to hand: \(p_{return}\).
d) Kinetic energy after 6 s: \(KE_{6}\).
e) Potential energy at highest point: \(PE_{max}\).
Equations:
- Velocity at time \(t\):
\[
v(t) = v_0 - g t.
\]
- Momentum:
\[
p(t) = m v(t).
\]
- Kinetic energy:
\[
KE = \frac{1}{2} m v^2.
\]
- Potential energy:
\[
PE = m g h,
\]
where \(h = v_0^2 / (2g)\) at the maximum height.
Calculations:
a) After 2 sec:
\[
v_{2} = 40 - 9.8 \times 2 = 40 - 19.6 = 20.4\, \text{m/s}.
\]
Assuming mass \(m = 0.16\, \text{kg}\):
\[
p_{2} = 0.16 \times 20.4 = 3.264\, \text{kg·m/s}.
\]
b) Highest point:
\[
v_{max} = 0\, \text{m/s},
\]
so
\[
p_{max} = 0.
\]
c) When it returns to hand:
\[
v_{return} = -v_0 = -40\, \text{m/s},
\]
so
\[
p_{return} = 0.16 \times (-40) = -6.4\, \text{kg·m/s}.
\]
d) Kinetic energy after 6 seconds:
\[
v_{6} = 40 - 9.8 \times 6 = 40 - 58.8 = -18.8\, \text{m/s},
\]
taking magnitude for KE calculation:
\[
KE = \frac{1}{2} \times 0.16 \times (18.8)^2 \approx 0.08 \times 353.44 \approx 28.27\, \text{J}.
\]
e) Potential energy at maximum height:
\[
h = \frac{v_0^2}{2g} = \frac{(40)^2}{2 \times 9.8} \approx \frac{1600}{19.6} \approx 81.63\, \text{m}.
\]
\[
PE_{max} = 0.16 \times 9.8 \times 81.63 \approx 0.16 \times 798.77 \approx 127.8\, \text{J}.
\]
Answer:
a) Momentum after 2 seconds: 3.264 kg·m/s.
b) Momentum at highest point: 0 kg·m/s.
c) Momentum when returning: -6.4 kg·m/s.
d) Kinetic energy after 6 seconds: 28.27 Joules.
e) Potential energy at maximum height: 127.8 Joules.
Problem 4: Force exerted when stopping a billiard ball
Known:
- Initial momentum: from previous problem \(p_{initial} = 3.264\, \text{kg·m/s}\).
- Time to stop, \(\Delta t = 1\, \text{s}\).
Unknown:
- Force exerted on hand, \(F\).
Equation:
- Impulse-momentum theorem:
\[
F \times \Delta t = \Delta p,
\]
which simplifies to
\[
F = \frac{\Delta p}{\Delta t}.
\]
Calculation:
\[
F = \frac{0 - 3.264}{1} = -3.264\, \text{N}.
\]
The negative sign indicates the force opposes movement.
Answer:
The magnitude of the force exerted on your hand is approximately 3.264 N.
Problem 5: Energy cost of lifting mass on planet X versus Earth
Known:
- Mass lifted: \(m = 10\, \text{kg}\).
- Height: \(h=2\, \text{m}\).
- Gravity on planet X: \(g_{X} = 36.8\, \text{m/s}^2\).
- Gravity on Earth: \(g_{e} = 9.8\, \text{m/s}^2\).
- Wage rate: $10 per joule.
Unknown:
- Work done on planet X: \(W_X\).
- Work on Earth: \(W_E\).
Equations:
- Work done:
\[
W = m g h.
\]
Calculations:
- On planet X:
\[
W_X = 10 \times 36.8 \times 2 = 736\, \text{J}.
\]
- On Earth:
\[
W_E = 10 \times 9.8 \times 2 = 196\, \text{J}.
\]
Wages:
- On planet X:
\[
\$ = 736 \times 10 = \$7360.
\]
- On Earth:
\[
\$ = 196 \times 10 = \$1960.
\]
Answer:
You make 7360 dollars on planet X and 1960 dollars on Earth for the same work at the same rate.
Problem 6: Power of a winch lifting a load
Known:
- Mass: \(m = 250\, \text{kg}\).
- Height: \(h = 20\, \text{m}\).
- Time: \(t = 5\, \text{seconds}\).
Unknown:
- Power output \(P\).
Equation:
- Power:
\[
P = \frac{W}{t} = \frac{m g h}{t}.
\]
Calculations:
\[
W = 250 \times 9.8 \times 20 = 49000\, \text{J}.
\]
\[
P = \frac{49000}{5} = 9800\, \text{W} = 9.8\, \text{kW}.
\]
Answer:
The power of the winch is 9.8 kW.
Problem 7: Stopping a truck moving at 70 mph in 20 meters
Known:
- Mass of truck: \(m = 1500\, \text{kg}\).
- Initial velocity: \(v_{i} = 70\, \text{mph} \approx 112.6\, \text{km/h} = 31.28\, \text{m/s}\).
- Stopping distance: \(d = 20\, \text{m}\).
Unknowns:
a) Force needed to stop each wheel.
b) Total work done.
Equations:
- Force from work-energy principle:
\[
W = \Delta KE = \frac{1}{2} m v_i^2,
\]
and
\[
F_{avg} = \frac{W}{d}.
\]
Calculations:
- Kinetic energy:
\[
KE = 0.5 \times 1500 \times (31.28)^2 \approx 0.75 \times 979.2 \approx 734.4\, \text{kJ}.
\]
- Work equals change in KE:
\[
W = 734,400\, \text{J}.
\]
- Force:
\[
F_{total} = \frac{W}{d} = \frac{734,400}{20} = 36,720\, \text{N}.
\]
- Force per wheel (assuming 4 wheels):
\[
F_{per\, wheel} = \frac{36,720}{4} = 9,180\, \text{N}.
\]
Answer:
a) Approximately 9,180 N applied to each wheel.
b) Total work done is approximately 734.4 kJ.
Problem 8: Using a lever to lift a heavy load
Known:
- Fulcrum distances: load \(d_{L} = 0.8\, \text{m}\), effort \(d_{E} = 2.2\, \text{m}\).
- Load mass: \(m_{L} = 200\, \text{kg}\).
- Gravitational acceleration: \(g = 9.8\, \text{m/s}^2\).
- Effort displacement: \(d_{effort} = 40\, \text{cm} = 0.4\, \text{m}\).
Unknowns:
a) Effort force \(F_{E}\).
b) Displacement of load \(d_{load}\).
Equations:
- Moment balance:
\[
F_{E} \times d_{E} = F_{L} \times d_{L},
\]
where \(F_{L} = m_{L} g\).
- Effort force:
\[
F_{E} = \frac{F_{L} \times d_{L}}{d_{E}}.
\]
- Displacement ratio:
\[
\frac{d_{load}}{d_{effort}} = \frac{d_{E}}{d_{L}}.
\]
Calculations:
a) Force:
\[
F_{L} = 200 \times 9.8 = 1960\, \text{N},
\]
\[
F_{E} = \frac{1960 \times 0.8}{2.2} \approx \frac{1568}{2.2} \approx 713\, \text{N}.
\]
b) Displacement:
\[
d_{load} = d_{effort} \times \frac{d_{L}}{d_{E}} = 0.4 \times \frac{0.8}{2.2} \approx 0.4 \times 0.364 = 0.146\, \text{m} \approx 14.6\, \text{cm}.
\]
Answer:
a) You need to apply approximately 713 N effort.
b) The load moves approximately 14.6 cm when effort moves 40 cm.
Problem 9: Seesaw balance between Donald and Hillary
Known:
- Donald's weight: \(W_D = 90\, \text{kg} \times 9.8 = 882\, \text{N}\).
- Hillary's weight: \(W_H = 65\, \text{kg} \times 9.8 = 637\, \text{N}\).
- Distance between Donald and Hillary: 3 m.
- Heights from fulcrum: \(d_D\) and \(d_H\).
Unknown:
- Distance from fulcrum for each person, assuming static equilibrium:
\[
W_D \times d_D = W_H \times d_H,
\]
and
\[
d_D + d_H = 3\, \text{m}.
\]
Calculations:
\[
d_D = \frac{W_H \times d_H}{W_D} = \frac{637 \times d_H}{882}.
\]
Since total length is 3m:
\[
d_D + d_H = 3 \Rightarrow \frac{637 \times d_H}{882} + d_H = 3,
\]
\[
d_H \left(\frac{637}{882} + 1\right) = 3,
\]
\[
d_H \left(\frac{637 + 882}{882}\right) = 3,
\]
\[
d_H \times \frac{1519}{882} = 3,
\]
\[
d_H = 3 \times \frac{882}{1519} \approx 3 \times 0.58 \approx 1.74\, \text{m}.
\]
\[
d_D = 3 - 1.74 = 1.26\, \text{m}.
\]
Answer:
- Donald should sit approximately 1.26 meters from the fulcrum.
- Hillary should sit approximately 1.74 meters from the fulcrum.
Problem 10: Mechanical advantage with pulleys
Given:
- Load: 100 N.
- Maximum effort: 100 N.
- Objective: Lift 10 metric tons (10,000 kg) (which is 98,100 N weight) by 1 cm.
a) Relationship:
- With ideal pulleys:
\[
\text{Mechanical advantage} \approx \frac{\text{Load}}{\text{Effort}}.
\]
- Displacement of effort relates to load displacement by the number of pulleys:
\[
\text{Effort displacement} = n \times \text{load displacement}.
\]
b) To lift 10,000 kg (98,100 N) with effort of 100 N:
\[
\text{Number of pulleys} = \frac{\text{Load}}{\text{Effort}} = \frac{98,100}{100} = 981.
\]
- Displacement of effort is:
\[
d_{effort} = \text{load displacement} \times n = 0.01\, \text{m} \times 981 \approx 9.81\, \text{m}.
\]
Answer:
a) The pulleys provide a mechanical advantage equal to the load divided by effort; to lift 10 tons with only 100 N effort, around 981 pulleys are needed. When lifting the load by 1 cm, the effort must be displaced about 9.81 meters.
Conclusion
These problems demonstrate the application of fundamental physics principles in scenarios involving motion, energy, work, levers, and pulleys. Understanding the relationship between known and unknown quantities, relevant