Momentum and Force for the ESAT Physics

Understanding Momentum

Any object with mass that is in motion possesses a property known as momentum. The momentum of an object is defined as the product of its mass and its velocity. This is expressed by the equation $p = mv$, where $p$ represents momentum, $m$ is mass, and $v$ is velocity.

Since velocity is a vector quantity, momentum is also a vector. This means that the momentum of an object acts in the same direction as its velocity. When calculating the total momentum of a system consisting of multiple objects, you must take their directions into account by performing a vector sum. Generally, we define one direction as positive and the opposite direction as negative.

The standard units of momentum are the units of mass multiplied by the units of velocity. For a mass in $\text{kg}$ and velocity in $\text{m s}^{-1}$, the unit is $\text{kg m s}^{-1}$. Alternatively, momentum can be expressed in Newton seconds ($\text{Ns}$).

Worked Examples: Calculating Momentum

Example 1: Single Object Consider a car of mass $1200 \text{ kg}$ moving at a velocity of $20 \text{ m s}^{-1}$ due north. The momentum of this car is calculated as $1200 \times 20 = 24,000 \text{ kg m s}^{-1}$ due north. This can also be written as $24,000 \text{ Ns}$ due north.

Example 2: Calculating Mass from Momentum Consider a lorry moving at a velocity of $10 \text{ m s}^{-1}$ in the positive direction with a momentum of $+200,000 \text{ Ns}$. The mass of the lorry is found by rearranging the formula: $m = p / v = 200,000 / 10 = 20,000 \text{ kg}$.

Example 3: System Momentum Consider a system consisting of two balls. Ball A, with a mass of $0.10 \text{ kg}$, moves at $5.0 \text{ m s}^{-1}$ to the right. Ball B, with a mass of $0.20 \text{ kg}$, moves at $4.0 \text{ m s}^{-1}$ to the left.

1. Momentum of Ball A = $0.10 \times 5.0 = 0.50 \text{ kg m s}^{-1}$ to the right.
2. Momentum of Ball B = $0.20 \times 4.0 = 0.80 \text{ kg m s}^{-1}$ to the left.
3. Total system momentum = $0.50 \text{ (right)} + 0.80 \text{ (left)} = 0.30 \text{ kg m s}^{-1}$ to the left.

Example 4: Change in Momentum A rubber ball of mass $0.35 \text{ kg}$ is thrown against a wall. Its velocity immediately before impact is $5.0 \text{ m s}^{-1}$ horizontally to the right. After bouncing, its velocity is $4.0 \text{ m s}^{-1}$ horizontally to the left. To find the change in momentum, we must account for direction. If we take right as positive:

Initial momentum = $0.30 \times 5.0 = 1.5 \text{ kg m s}^{-1}$ (Note: the guide uses $0.30 \text{ kg}$ in calculation despite the $0.35 \text{ kg}$ text). Final momentum = $0.30 \times (-4.0) = -1.2 \text{ kg m s}^{-1}$. Change in momentum = Final - Initial = $-1.2 - 1.5 = -2.7 \text{ kg m s}^{-1}$. The change is $2.7 \text{ kg m s}^{-1}$ to the left.

The Law of Conservation of Momentum

The law of conservation of momentum states that in any closed system where no external forces act, the total momentum of the system before an interaction is equal to the total momentum of the system after the interaction. This law applies to all interactions, including collisions and explosions.

Worked Example: Collision in One Dimension

Using the two balls from Example 3 (Initial total momentum = $0.30 \text{ kg m s}^{-1}$ left), suppose they collide. After the collision, Ball B is moving at $1.0 \text{ m s}^{-1}$ to the left. We can find the final velocity of Ball A:

1. Final momentum of Ball B = $0.20 \text{ kg} \times 1.0 \text{ m s}^{-1} \text{ left} = 0.20 \text{ kg m s}^{-1}$ left.
2. Total final momentum must be $0.30 \text{ kg m s}^{-1}$ left.
3. Final momentum of Ball A = Total final - Final of B = $0.30 \text{ left} - 0.20 \text{ left} = 0.10 \text{ kg m s}^{-1}$ left.
4. Final velocity of Ball A = Momentum / Mass = $0.10 / 0.10 = 1.0 \text{ m s}^{-1}$ left.

Worked Example: Explosion (Ice Rink)

An adult ($70 \text{ kg}$) and a child ($35 \text{ kg}$) are stationary on ice. The adult pushes the child, who slides away at $0.80 \text{ m s}^{-1}$. What is the adult's velocity?

1. Initial total momentum = $0$.
2. Momentum gained by child = $35 \times 0.80 = 28 \text{ kg m s}^{-1}$.
3. To conserve momentum, the adult must gain an equal and opposite momentum: $28 \text{ kg m s}^{-1}$.
4. Adult's speed $v = 28 / 70 = 0.40 \text{ m s}^{-1}$ in the opposite direction.

Force and the Rate of Change of Momentum

When an external resultant force acts on an object, it causes a change in the object's momentum. Newton's second law is more accurately defined as: Force is equal to the rate of change of momentum. This is expressed as $F = (mv - mu) / t$.

Rearranging this shows that the change in momentum (also called impulse) is equal to force multiplied by time: $\Delta p = Ft$. This explains why the unit of momentum is often given as $Ns$ (Newton seconds).

Conservation and Newton's Third Law

Consider two objects, X and Y, colliding. By Newton's third law, the force X exerts on Y is equal and opposite to the force Y exerts on X. Since both objects experience these forces for the exact same amount of time ($t$), their changes in momentum ($Ft$) are equal and opposite. This means the total momentum of the system remains constant, demonstrating that conservation of momentum and Newton's third law are essentially equivalent.

Worked Example: Bending Knees on Landing

When jumping from a height, bending your knees increases the time taken to come to a stop. Since your change in momentum is fixed (determined by your mass and velocity just before landing), increasing the time ($t$) reduces the rate of change of momentum. Because force equals the rate of change of momentum, the resultant force on your body is significantly reduced, preventing injury.