AP Physics Unit

Unit 1 - Vectors and Kinematics

Intermediate

Mathematical

FRQ

A rocket is sent to shoot down an invading spacecraft that is hovering at an altitude of 1500 meters. The rocket is launched with an initial velocity of 180 m/s. Find the following:

  1. (a) The minimum velocity needed to hit the spaceship. (3 points)
  2. (b) The time taken for the rocket to reach its maximum height when launched at 180 m/s. (2 points)
  3. (c) The maximum height the rocket reaches when launched at 180 m/s. (2 points)
  4. (d) The velocity and time of impact (assuming that the rocket collides with the spacecraft). (5 points)
  1. Minimum Velocity Needed to Hit the Spaceship:
    • v_{\text{min}} \approx 171.46 , \text{m/s}
  2. Time Taken to Reach Maximum Height at 180 m/s:
    • t \approx 18.37 , \text{s}
  3. Maximum Height Reached at 180 m/s:
    • h_{\text{max}} \approx 1653.06 , \text{m}
  4. Velocity and Time of Impact with the Spacecraft:
    • Velocity at Impact: v \approx 54.77 , \text{m/s}
    • Time of Impact: t \approx 12.78 , \text{s}
0
  1. Minimum Velocity Needed to Hit the Spaceship:

The minimum velocity is the initial vertical velocity required to reach 1500 m under the influence of gravity.

Step Formula Derivation Reasoning
1.1 v_{\text{min}} = \sqrt{2gh} The minimum velocity required, derived from the equation v^2 = v_0^2 + 2gh, where v_0 is initial velocity and v is final velocity (0 at the top).
  1. Time to Reach Maximum Height at 180 m/s:

This is the time taken for the rocket to reach its highest point, where its vertical velocity becomes zero.

Step Formula Derivation Reasoning
2.1 t = \frac{v_0}{g} Time to reach maximum height, where v_0 is initial vertical velocity and g is acceleration due to gravity.
  1. Maximum Height Reached at 180 m/s:

The highest point the rocket reaches when launched vertically at 180 m/s.

Step Formula Derivation Reasoning
3.1 h = \frac{v_0^2}{2g} Maximum height reached, derived from v^2 = v_0^2 – 2gh, with v = 0 at the maximum height.
  1. Velocity and Time of Impact with the Spacecraft:

Assuming the rocket reaches the spacecraft, we calculate the time and velocity of impact.

Step Formula Derivation Reasoning
4.1 v = \sqrt{v_0^2 – 2gh} Velocity at impact, derived from v^2 = v_0^2 – 2gh.
4.2 t = \frac{v_0 – v}{g} Time of impact, calculated from v = v_0 – gt.

We will now calculate these values.

Here are the results:

  1. Minimum Velocity Needed to Hit the Spaceship:
    • v_{\text{min}} \approx 171.46 , \text{m/s}
  2. Time Taken to Reach Maximum Height at 180 m/s:
    • t \approx 18.37 , \text{s}
  3. Maximum Height Reached at 180 m/s:
    • h_{\text{max}} \approx 1653.06 , \text{m}
  4. Velocity and Time of Impact with the Spacecraft:
    • Velocity at Impact: v \approx 54.77 , \text{m/s}
    • Time of Impact: t \approx 12.78 , \text{s}

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  1. Minimum Velocity Needed to Hit the Spaceship:
    • v_{\text{min}} \approx 171.46 , \text{m/s}
  2. Time Taken to Reach Maximum Height at 180 m/s:
    • t \approx 18.37 , \text{s}
  3. Maximum Height Reached at 180 m/s:
    • h_{\text{max}} \approx 1653.06 , \text{m}
  4. Velocity and Time of Impact with the Spacecraft:
    • Velocity at Impact: v \approx 54.77 , \text{m/s}
    • Time of Impact: t \approx 12.78 , \text{s}

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: \text{5 km}

  2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

  3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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