Momentum and Force Relations

Step 3: Calculate the final answer with units.

• $p = 50,000\ kg\ m/s$

Conservation of Momentum

The Law of Conservation of Momentum states that the total momentum before an event is exactly equal to the total momentum after the event.

This rule only applies in a closed system, which is defined as a system that has no external forces (like friction or air resistance) acting upon it.

Worked Example: Inelastic Collision

A 1200 kg car travelling at 15 m/s crashes into a stationary 800 kg car. They lock together upon impact. Calculate their combined velocity immediately after the collision.

Step 1: Calculate the total momentum before the collision.

• $p_{before} = (1200 \times 15) + (800 \times 0)$
• $p_{before} = 18,000 + 0 = 18,000\ kg\ m/s$

Step 2: Apply the Law of Conservation of Momentum.

• $Total\ p_{after} = Total\ p_{before} = 18,000\ kg\ m/s$
• $Total\ mass = 1200 + 800 = 2000\ kg$

Step 3: Rearrange the formula to find the final velocity.

• $v = \frac{p}{m}$
• $v = \frac{18,000}{2000}$
• $v = 9.0\ m/s$

Worked Example: Explosion and Recoil

A stationary 3.0 kg rifle fires a 0.02 kg bullet at a velocity of 300 m/s. Calculate the recoil velocity of the rifle.

Step 1: State the momentum before the explosion.

• Both objects are initially stationary, so $Total\ p_{before} = 0\ kg\ m/s$.

Step 2: Apply conservation of momentum to the system after firing.

• $0 = (m_{bullet} \times v_{bullet}) + (m_{rifle} \times v_{rifle})$
• $0 = (0.02 \times 300) + (3.0 \times v_{recoil})$

Step 3: Calculate the recoil velocity.

• $0 = 6.0 + 3.0v_{recoil}$
• $-6.0 = 3.0v_{recoil}$
• $v_{recoil} = -2.0\ m/s$ (The negative sign shows the rifle moves in the opposite direction to the bullet).

Force and the Rate of Change of Momentum

Why does dropping a glass on a concrete floor smash it, but dropping it on a thick carpet from the same height leaves it intact? The answer lies in how quickly the glass comes to a halt.

Newton's Second Law states that the resultant force acting on an object is equal to its rate of change of momentum. This means that force depends on how much the momentum changes, divided by the time it takes to change.

The impulse is the product of force and time, and it is directly equal to the change in momentum. The relationship between force, mass, velocity, and time is given by:

$$F = \frac{mv - mu}{t}$$

Where:

• $F$ = resultant force ($N$)
• $m$ = mass ($kg$)
• $v$ = final velocity ($m/s$)
• $u$ = initial velocity ($m/s$)
• $t$ = time taken ($s$)

Deriving the Momentum Equation

In the exam, you must be able to explain how the momentum-force equation connects to standard acceleration. We start with the definition of inertial mass via Newton's Second Law:

Step 1: State Newton's Second Law.

• $F = m \times a$

Step 2: Substitute the formula for acceleration ($a = \frac{v - u}{t}$).

• $F = m \times \frac{v - u}{t}$

Step 3: Expand the bracket.

• $F = \frac{mv - mu}{t}$

Note that $mv - mu$ represents the total change in momentum ($\Delta p$), proving that $F = \frac{\Delta p}{t}$.

Worked Example: Rebound Force

A 0.4 kg football hits a wall at 12 m/s and bounces back directly at 10 m/s. The impact with the wall lasts for 0.2 s. Calculate the resultant force exerted by the wall on the ball.

Step 1: Identify the initial and final velocities, remembering direction.

• Let motion towards the wall be positive: $u = 12\ m/s$.
• The ball rebounds in the opposite direction: $v = -10\ m/s$.

Step 2: Calculate the change in momentum.

• $\Delta p = m(v - u)$
• $\Delta p = 0.4 \times (-10 - 12)$
• $\Delta p = 0.4 \times (-22) = -8.8\ kg\ m/s$

Step 3: Calculate the resultant force.

• $F = \frac{\Delta p}{t}$
• $F = \frac{-8.8}{0.2} = -44\ N$ (The negative sign indicates the force acts in the opposite direction to the initial movement).

Applying Momentum to Road Safety

Every time you step into a car, you are surrounded by engineering designed to manipulate the physics of momentum. When a crash occurs, the passengers must undergo a specific fixed change in momentum to come to a complete stop.

Safety features like crumple zones, airbags, and seatbelts all share the same fundamental purpose: they are designed to increase the time taken for the collision to occur. Seatbelts stretch slightly, crumple zones deform, and airbags compress.

Because force is inversely proportional to impact time ($F \propto \frac{1}{t}$), increasing the time significantly decreases the rate of change of momentum. This drastically reduces the resultant force experienced by the passengers, lowering the risk of severe injury.

Worked Example: The Impact of Safety Features

An 80 kg driver travelling at 25 m/s is involved in a collision and comes to a complete stop. Compare the force acting on the driver if they stop in 0.1 s (hitting the dashboard) versus 0.8 s (deploying an airbag).

Step 1: Calculate the fixed change in momentum for the driver.

• $\Delta p = m(v - u)$
• $\Delta p = 80 \times (0 - 25) = -2000\ kg\ m/s$

Step 2: Calculate the force for Case A (Dashboard, $t = 0.1\ s$).

• $F = \frac{-2000}{0.1} = -20,000\ N$

Step 3: Calculate the force for Case B (Airbag, $t = 0.8\ s$).

• $F = \frac{-2000}{0.8} = -2500\ N$

Step 4: State your conclusion.

• Increasing the impact time by a factor of 8 reduces the resultant force by a factor of 8, making the crash much more survivable.