AS Level Physics - Dynamics Flashcards

Flashcards testing core concepts, calculations, and derivations from AS Level Physics Dynamics notes including impulse, momentum, collisions, pulley systems, recoil, and Newton's laws.

What is the definition of Impulse?

Impulse is defined as the product of the force acting on an object and the time for which it acts.

What are the formulas and SI unit for Impulse ($I$)?

The formulas are $I = F \times t$, $I = \frac{\text{d}p}{\text{d}t}$, $I = \text{d}p$, $I = p_f - p_i$, and $I = mv - mu$. The unit of impulse is $\text{N}\text{s}$.

What physical quantity is represented by the area under a force ($F$) versus time ($t$) graph?

The area under a force-time graph represents the product of $F$ and $t$, which gives the impulse or change in momentum ($\text{d}p$).

What is the change in momentum ($\text{d}p$) when a ball of mass $m$ collides elastically with a wall at velocity $v$ and rebounds?

Taking the rebound direction as positive, $\text{d}p = p_f - p_i = mv - (-mv) = 2mv$ (or $-2mv$ if the rebound direction is taken as negative).

A ball of mass $m = 0.05\text{ kg}$ hits the ground at $-14\text{ m/s}$ and rebounds at $+6\text{ m/s}$ in an impact lasting $t = 0.01\text{ s}$. What is the change in momentum and the force during impact?

The change in momentum is $\text{d}p = 6(0.05) - (0.05)(-14) = 1\text{ N}\text{s}$. The force during impact is $F = \frac{\text{d}p}{t} = \frac{1}{0.01} = 100\text{ N}$.

In what direction is tension force always directed?

Tension is always directed away from the concerned object.

What is the tension $T$ in a rope pulling a $2\text{ kg}$ object vertically upward at a constant velocity of $6\text{ m/s}$ (assuming $g = 10\text{ m/s}^2$)?

Since velocity is constant, forces are balanced ($F_{\text{net}} = 0$), so $T = W = 20\text{ N}$.

What is the tension $T$ in a rope pulling a $2\text{ kg}$ object vertically upward with an acceleration of $2\text{ m/s}^2$ (assuming $g = 10\text{ m/s}^2$)?

Using $F = ma$, $T - 20 = 2(2)$, which gives $T = 24\text{ N}$.

How is a downward deceleration of $5\text{ m/s}^2$ treated when applying $F = ma$?

A downward deceleration is equivalent to an upward acceleration of $5\text{ m/s}^2$.

A driving force of $2000\text{ N}$ acts on a $20\text{ kg}$ object. What is the resistive force $F_R$ if the object decelerates at $6\text{ m/s}^2$?

Using $F = ma$, $2000 - F_R = 20(-6)$, which gives $F_R = 2120\text{ N}$.

Two masses $A = 3\text{ kg}$ and $B = 2\text{ kg}$ are connected over a frictionless pulley. What is the acceleration $a$ and tension $T$ in the string (assuming $g = 10\text{ m/s}^2$)?

Applying $F = ma$ separately: $30 - T = 3a$ and $T - 20 = 2a$. Solving simultaneously gives $a = 2\text{ m/s}^2$ and $T = 24\text{ N}$.

State the Principle of Conservation of Momentum.

The total momentum of a system always remains constant or conserved, provided that there is no external force acting on the system.

What are the four defining properties of an elastic collision?

1. Momentum is conserved. 2. Kinetic Energy (KE) is conserved. 3. Total Energy (TE) is conserved. 4. Speed of Approach (S.O.A) equals Speed of Separation (S.O.S).

What are the defining properties of an inelastic collision?

1. Momentum is conserved. 2. Total Energy (TE) is conserved. 3. Kinetic Energy (KE) is NOT conserved. 4. Speed of Approach does not equal Speed of Separation (S.O.A $\neq$ S.O.S).

How are Speed of Approach (S.O.A) and Speed of Separation (S.O.S) calculated for bodies moving in opposite directions versus the same direction?

For bodies moving in opposite directions, relative speeds are added. For bodies moving in the same direction, relative speeds are subtracted (larger speed minus smaller speed).

What can be immediately concluded if two colliding particles stick together and move with a common velocity?

The collision can be directly concluded to be inelastic because kinetic energy is lost.

When applying the Principle of Conservation of Momentum to two-dimensional collisions, how are the equations set up?

Momentum is resolved into components and conserved separately in two perpendicular directions: horizontally ($p_x$ before = $p_x$ after) and vertically ($p_y$ before = $p_y$ after).

Why does a gun recoil backward when a bullet is fired from rest?

Since the initial momentum of the system is zero, the final total momentum must remain zero. The forward momentum of the bullet must be cancelled by an equal and opposite backward momentum of the gun.

What is the relationship between mass and velocity for two bodies separating from rest under zero external force?

Mass and velocity are inversely proportional ($M \propto \frac{1}{V}$ or $MV = mv$). An object with $400$ times the mass will recoil with $400$ times less velocity.

What is the relationship between the ratio of kinetic energies and the ratio of masses for a gun and bullet system fired from rest?

$\frac{\text{KE}_{\text{Gun}}}{\text{KE}_{\text{Bullet}}} = \frac{v_{\text{Gun}}}{v_{\text{Bullet}}} = \frac{m_{\text{Bullet}}}{m_{\text{Gun}}}$.

What is the value of 1 atomic mass unit ($u$) in kilograms?

$1\text{ u} = 1.66 \times 10^{-27}\text{ kg}$.

A Radon nucleus ($M_R = 220$) at rest decays into a Radium nucleus ($M_R = 216$) and a Helium nucleus ($M_R = 4$). If the Helium nucleus speed is $5.5 \times 10^5\text{ m/s}$, what is the recoil velocity $n$ of the Radium nucleus?

Applying conservation of momentum: $0 = (4u)(5.5 \times 10^5) + (216u)(-n)$, which gives recoil velocity $n = 1.0 \times 10^3\text{ m/s}$.

How is Kinetic Energy (KE) expressed in terms of momentum ($p$) and mass ($m$)?

Since $\text{KE} = \frac{1}{2}mv^2$ and $p = mv$ (or $v = \frac{p}{m}$), substituting $v$ yields $\text{KE} = \frac{p^2}{2m}$.

What three main conditions characterize Newton's Third Law of Motion?

1. The two forces are equal in magnitude. 2. The two forces act in opposite directions. 3. The two forces act on different bodies ($F_A = -F_B$).

What is the value of the contact action/reaction force between two bodies in free fall?

During free fall, the action/reaction contact forces between the falling objects are always zero ($n = 0\text{ N}$).

For an object of mass $m$ on a surface inclined at angle $\theta$, what are the parallel and perpendicular components of weight?

The component of weight parallel to the inclined surface is $mg \sin(\theta)$, and the component perpendicular to the surface is $mg \cos(\theta)$.