Physics Test: Hooke's Law, Forces, Friction, and Momentum

Comprehensive practice flashcards covering Hooke's Law, free body diagrams, friction, air resistance, and momentum calculations from the Physics test.

According to Hooke's Law, what does the graph of force versus extension look like before the limit of proportionality is reached?

A straight line passing through the origin, indicating that extension is directly proportional to the applied force.

Which modification will increase the extension of a hanging spring?

Using a heavier mass, which exerts a greater downward force on the spring.

In a free body diagram of a box resting on a table, which upward force balances the downward force of gravity ($mg$)?

The normal force.

In which direction does friction always act relative to the motion of an object?

Friction acts opposite to the direction of motion.

If a car is moving to the right along a surface, in which direction does the friction force act?

To the left (opposite to the direction of motion).

Which factors primarily affect the air resistance experienced by a moving object?

Its surface area and shape.

If two parachutes carrying the same load are dropped from the same height, which one reaches the ground last?

The parachute with the large canopy reaches the ground last.

Why does a parachute with a larger canopy take longer to fall than one with a smaller canopy carrying the same load?

The larger canopy has a greater surface area, which produces more air resistance (drag), leading to a lower terminal velocity.

If two objects have identical momentum, what must be true about them?

The product of their mass and velocity ($p = m \times v$) is the same.

A loaded airport baggage trolley and an empty trolley travel at the same velocity. Which trolley has more momentum, and why?

The loaded trolley has more momentum because it has a greater mass, and momentum is the product of mass and velocity ($p = m \times v$).

Between an object of $0.5\,\text{kg}$ at $1\,\text{m/s}$, $1\,\text{kg}$ at $1\,\text{m/s}$, $1\,\text{kg}$ at $3\,\text{m/s}$, and $2\,\text{kg}$ at $1\,\text{m/s}$, which has the greatest momentum?

The $1\,\text{kg}$ object moving at $3\,\text{m/s}$ has the greatest momentum ($p = 1\,\text{kg} \times 3\,\text{m/s} = 3\,\text{kg}\cdot\text{m/s}$).

How does increasing the mass hanging from a spring affect its extension, according to Hooke's Law?

Increasing the mass increases the downward gravitational force (weight, $F = mg$). According to Hooke's Law ($F \propto x$), the extension increases directly proportionally to this applied force.

What are the two vertical forces acting on a box resting stationary on a flat table in a free body diagram?

The downward force of gravity (weight, $mg$) and the upward normal contact force from the table.

Trolley A has a mass of $4\,\text{kg}$ and moves at $3\,\text{m/s}$. What is the momentum of Trolley A?

$p = m \times v = 4\,\text{kg} \times 3\,\text{m/s} = 12\,\text{kg}\cdot\text{m/s}$.

Trolley B has a mass of $2\,\text{kg}$ and moves at $5\,\text{m/s}$. What is the momentum of Trolley B?

$p = m \times v = 2\,\text{kg} \times 5\,\text{m/s} = 10\,\text{kg}\cdot\text{m/s}$.

A student states: 'The faster trolley always has greater momentum.' Is this statement correct based on Trolley A ($4\,\text{kg}$, $3\,\text{m/s}$) and Trolley B ($2\,\text{kg}$, $5\,\text{m/s}$)?

No, the statement is incorrect. Trolley B is faster ($5\,\text{m/s} > 3\,\text{m/s}$), but Trolley A has greater momentum ($12\,\text{kg}\cdot\text{m/s} > 10\,\text{kg}\cdot\text{m/s}$) because momentum depends on both mass and velocity.

A $3\,\text{kg}$ object moves at $4\,\text{m/s}$. What is its initial momentum?

$p = m \times v = 3\,\text{kg} \times 4\,\text{m/s} = 12\,\text{kg}\cdot\text{m/s}$.

A $3\,\text{kg}$ object moves at $4\,\text{m/s}$. If its velocity doubles to $8\,\text{m/s}$, what is its new momentum?

$p = m \times v = 3\,\text{kg} \times 8\,\text{m/s} = 24\,\text{kg}\cdot\text{m/s}$.

An object's velocity doubles from $4\,\text{m/s}$ to $8\,\text{m/s}$. How must its mass change to keep its original momentum of $12\,\text{kg}\cdot\text{m/s}$?

Its mass must be halved, reducing from $3\,\text{kg}$ to $1.5\,\text{kg}$ ($m = \frac{p}{v} = \frac{12\,\text{kg}\cdot\text{m/s}}{8\,\text{m/s}} = 1.5\,\text{kg}$).