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**Part (a): Find the speed of the third piece**

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

1 | Let [katex] m [/katex] be the mass of each smaller piece, thus the mass of the third piece is [katex] 2.5m [/katex]. | According to the problem, the third piece has 2.5 times the mass of each of the other two pieces. |

2 | [katex] \vec{p}_{\text{total}} = \vec{0} [/katex] | The total initial momentum is zero since the coconut was stationary before exploding. |

3 | Let [katex] \vec{v}_3 [/katex] be the velocity of the third piece and [katex] \theta [/katex] its angle from west towards south. Then, [katex] \vec{p}_{\text{total}} = m \vec{v}_S + m \vec{v}_W + 2.5m \vec{v}_3 = \vec{0} [/katex] [katex] \vec{v}_S = 18\, \text{m/s} \, \hat{j} \quad \text{and} \quad \vec{v}_W = -18\, \text{m/s} \, \hat{i} [/katex] [katex] -m \cdot 18 \, \hat{i} + m \cdot 18 \, \hat{j} + 2.5m \vec{v}_3 = \vec{0} [/katex] |
Set the total momentum as the vector sum of individual momenta. The pieces are moving south and west with the same speed but in perpendicular directions. |

4 | [katex] \vec{v}_3 = \left(\frac{18}{2.5}\right) \hat{i} – \left(\frac{18}{2.5}\right) \hat{j} [/katex] [katex] \vec{v}_3 = 7.2 \hat{i} – 7.2 \hat{j} \; \text{m/s} [/katex] |
Rearrange to find the velocity vector of the third piece. Cancel [katex]m[/katex] and solve for [katex] \vec{v}_3 [/katex]. |

5 | [katex] \text{Speed of third piece } |\vec{v}_3| = \sqrt{(7.2)^2 + (7.2)^2} = \sqrt{103.68} \approx 10.18 \; \text{m/s} [/katex] |
Calculate the magnitude to find the speed of the third piece. |

6 | 10.18 m/s |
Answer for part (a), the speed of the third piece. |

**Part (b): Find the direction of the third piece**

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

1 | [katex] \tan(\theta) = \frac{-7.2}{7.2} = -1 [/katex] [katex] \theta = \tan^{-1}(-1) = 135^\circ \, \text{(from east counterclockwise)} [/katex] |
The angle [katex] \theta [/katex] is measured from the negative x-axis, thus the piece is moving to the northeast. |

2 | 135 degrees from east or northwest |
Answer for part (b). |

**Part (c): Reducing the impact force of the collision for the bystander**

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

1 | [katex] F = \frac{\Delta p}{\Delta t} [/katex] | Force experienced by the bystander can be reduced by increasing the impact time [katex]\Delta t[/katex] or reducing the momentum change [katex]\Delta p[/katex]. |

2 | Wear protective gear or position a net/barrier | By wearing protective gear or positioning an absorbent barrier (like a net), the bystander can prolong the impact time and reduce the force. |

3 | Use protective measures or barriers |
Answer for part (c), suggesting that protective gear or an impact-absorbing barrier would reduce the effect of the collision. |

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- Statistics

Intermediate

Mathematical

FRQ

A 70 kg woman and her 35 kg son are standing at rest on an ice rink, as shown above. They push against each other for a time of 0.60 s, causing them to glide apart. The speed of the woman immediately after they separate is 0.55 m/s.

Assume that during the push, friction is negligible compared with the forces the people exert on each other.

- Calculate the initial speed of the son after the push.
- Calculate the magnitude of the average force exerted on the son by the mother during the push.
- How do the magnitude and direction of the average force exerted on the mother by the son during the push compare with those of the average force exerted on the son by the mother? Justify your answer.
- After the initial push, the friction that the ice exerts cannot be considered negligible, and the mother comes to rest after moving a distance of 7.0 m across the ice. If their coefficients of friction are the same, how far does the son move after the push?

- Momentum

Advanced

Mathematical

GQ

A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?

- 1D Kinematics, Energy, Linear Forces, 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 truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation?

- Momentum

Beginner

Mathematical

MCQ

A 1200-kg car moving at 15.6 m/s suddenly collides with a stationary car of mass 1500 kg. If the two vehicles lock together, what is their combined velocity immediately after the collision?

- Momentum

- 10.18 m/s
- 135 degrees from east or northwest
- Use protective measures or barriers that increase the time of impact, and therefore reduce the force imparted by the mass.

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

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

[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |

[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |

[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |

Circular Motion | Energy |
---|---|

[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |

[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |

[katex]KE_i + PE_i = KE_f + PE_f[/katex] |

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

[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |

[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |

[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |

Simple Harmonic Motion |
---|

[katex]F = -k x[/katex] |

[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |

[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |

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

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

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

[katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |

[katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |

[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |

[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |

[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |

Variable | SI Unit |
---|---|

[katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |

[katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |

[katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |

[katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |

[katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |

Variable | Derived SI Unit |
---|---|

[katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |

[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |

[katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |

[katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |

[katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |

[katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |

[katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |

[katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |

General Metric Conversion Chart

Conversion Example

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

Start with the given measurement:

`[katex]\text{5 km}[/katex]`

Use the conversion factors for kilometers to meters and meters to millimeters:

`[katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]`

Perform the multiplication:

`[katex]\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}[/katex]`

Simplify to get the final answer:

`[katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]`

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

Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 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 [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
- 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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