An object at rest suddenly explodes into two fragments (m_{1} and m_{2}) by an explosion. Fragment m_{1} acquires 3 times the kinetic energy of the other. What is the ratio of m_{1} to m_{2}?

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

1 | KE_1 = 3KE_2 | Given: Kinetic energy of fragment 1 (KE_1) is 3 times that of fragment 2 (KE_2). |

2 | KE = \frac{1}{2}mv^2 | Kinetic energy formula. |

3 | \frac{1}{2}m_1v_1^2 = 3 \times \frac{1}{2}m_2v_2^2 | Substitute kinetic energy expressions for both fragments. |

4 | m_1v_1^2 = 3m_2v_2^2 | Simplify the equation. |

5 | m_1v_1 = m_2v_2 | Conservation of momentum (initial momentum is zero, so final momenta of fragments must be equal and opposite). |

6 | v_2 = \frac{m_1}{m_2}v_1 | Solve for v_2 from the momentum equation. |

7 | m_1v_1^2 = 3m_2\left(\frac{m_1}{m_2}v_1\right)^2 | Substitute v_2 into the kinetic energy equation. |

8 | m_1 = 3m_2\left(\frac{m_1^2}{m_2^2}\right) | Simplify the equation. |

9 | m_2 = 3m_1 | Rearrange the equation to solve for the ratio of masses. |

10 | m_1 /m_2 = \frac{1}{3} | Solve for the ratio of m_1 to m_2. |

11 | m_1:m_2 = 1/3 | Corrected ratio of the masses of the two fragments. |

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Ask Phy About This MCQ

Phy Beta V5 (1.28.24) – Systems Operational.

- Statistics

A block of mass m is moving on a horizontal frictionless surface with a speed v_0 as it approaches a block of mass 2m which is at rest and has an ideal spring attached to one side.

When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.

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?

The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 0.050 kg.

Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored.

When answering the following questions, consider the marble when it is at point C.

^{3} N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

^{2}. How far up does it move?

_{p} = 1.67 x10^{-27} kg) is being accelerated along a straight line at 3.6 ×10^{15} m/s^{2} in a machine. The proton has an initial speed of 2.4 x10^{7} m/s and travels 3.5 cm.

*v* by doing 5.0 x 10^{3} J of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?

^{2} and is lifted through a distance of 11 m.

A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.

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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!