Step | Derivation/Formula | Reasoning |
---|---|---|

1 | KE_{\text{initial}} = \frac{1}{2}m_{b}v_0^2 | Initial kinetic energy of the bullet, where m_b is the mass of the bullet and v_0 is the velocity of the bullet. |

2 | p_{\text{initial}} = m_b v_0 | Initial momentum of the bullet, calculated using momentum formula, p = mv . |

3 | p_{\text{after-collision}} = (m_b + m_w) v | After the collision, using the conservation of momentum, the total system (bullet + block) momentum is equal to the initial momentum. |

4 | m_b v_0 = (m_b + m_w) v | This step shows the conservation of momentum applied to the system. Solving for v gives the speed immediately after the collision. |

5 | KE_{\text{after-collision}} = \frac{1}{2}(m_b + m_w)v^2 | The kinetic energy of the bullet and block system immediately after the collision. |

6 | PE_{\text{max}} = m_gh | Potential energy when the block and bullet are at the maximum height h , with m_g = (m_b + m_w)g being the combined mass times the acceleration due to gravity. |

7 | KE_{\text{after-collision}} = PE_{\text{max}} | According to the conservation of mechanical energy, assuming no other forces like air resistance are acting, the kinetic energy immediately after collision equals the potential energy at the maximum height. |

8 | m_g h = \frac{1}{2} (m_b + m_w) v^2 | Simplifying above equations to match potential and kinetic energies. |

9 | Analyze statement (e) | Careful analyzation shows that the initial kinetic energy of the bullet is converted through an inelastic collision and ultimately equals the potential energy at the maximum height, assuming no energy lost to heat, etc. |

10 | Conclusion | Statement (e) “The initial kinetic energy of the bullet before the collision is equal to the potential energy of the bullet and block when they reach the maximum height h ” is correct. |

Phy can also check your working. Just snap a picture!

- Statistics

Advanced

Mathematical

GQ

A 1000 kg car is traveling east at 20m/s when it collides perfectly inelastically with a northbound 2000 kg car traveling at 15m/s. If the coefficient of kinetic friction is 0.9, how far, and at what angle do the two cars skid before coming to a stop?

- 1D Kinematics, Friction, Momentum

Beginner

Mathematical

GQ

Car A, mass 1000 kg, is traveling at 40 m/s when it collides with a stationary car B. They stick together and travel at 7 m/s. What is the mass of car B?

- Momentum

Intermediate

Mathematical

MCQ

A golf club exerts an average horizontal force of 1000 N on a 0.045-kg golf ball that is initially at rest on the tee. The club is in contact with the ball for 1.8 milliseconds. What is the speed of the golf ball just as it leaves the tee?

- Impulse, Momentum

Intermediate

Conceptual

MCQ

Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance d. Ignore friction and assume that an equal force *F* is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?

- Energy

Beginner

Mathematical

MCQ

A block with mass m slides at speed v_0 on a smooth surface and hits a stationary block with mass M . They stick together and move at speed v_0/3 . Find M in terms of m .

- Momentum

Intermediate

Mathematical

FRQ

Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.

- Energy, Momentum

Intermediate

Conceptual

MCQ

A rubber ball bounces off of a wall with an initial speed v and reverses its direction so its speed is v right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?

- Energy, Momentum

Advanced

Mathematical

GQ

A 2kg object slides east at 4 m/s and collides with a stationary 3 kg object. After the collision, the 2 kg object is traveling at an unknown velocity at 15° north of east and the 3 kg object is traveling at 38° south of east. What is each object’s final velocity?

- Momentum

Intermediate

Mathematical

MCQ

A force *F* is exerted by a broom handle on the head of a broom, which has a mass *m*. The handle is at an angle θ to the horizontal. The work done by the force on the head of the broom as it moves a distance d across a horizontal floor is:

- Energy

Intermediate

Mathematical

FRQ

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 18 m/s. The third piece has 2.5 times the mass as the other two.

- Momentum

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Kinematics | Forces |
---|---|

\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |

v = v_i + at | F_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 Motion | Energy |
---|---|

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 |

Momentum | Torque and Rotations |
---|---|

p = m v | \tau = r \cdot F \cdot \sin(\theta) |

J = \Delta p | I = \sum mr^2 |

p_i = p_f | L = I \cdot \omega |

Simple Harmonic Motion |
---|

F = -k x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients 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. |

k | Spring 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 |

Variable | SI 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)} |

Variable | Derived 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

Conversion Example

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

Start with the given measurement:

`\text{5 km}`

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}}`

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}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - 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 .
- Bookmark questions that you can’t solve so you can come back to them later.
- 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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