Question

A \(2000-\mathrm{kg}\) truck traveling north at \(40.0 \mathrm{~km} / \mathrm{h}\) turns east and accelerates to \(50.0 \mathrm{~km} / \mathrm{h}\). What is the magnitude and direction of the change of the truck's momentum?

Short answer

The magnitude of the change in the truck's momentum is 35507.22 kg.m/s, and the direction is 38.66 degrees north of east.

Step-by-step solution

Conversion of velocities

Convert the velocities from kilometers per hour to meters per second as it's the SI unit of velocity. For this, multiply the given velocities by \( \frac{1000}{3600} \). So, \( v1 = (40 \frac{km}{h})(\frac{1000 m}{1 km})(\frac{1 h}{3600 s}) = 11.11 \frac{m}{s} \) and \( v2 = (50 \frac{km}{h})(\frac{1000 m}{1 km})(\frac{1 h}{3600 s}) = 13.89 \frac{m}{s} \).

Calculate initial and final momenta

Momentum (p) is the product of mass (m) and velocity (v). So, \( p = mv \). Initial momentum, \( p_{i} \), when the truck was moving north: \( p_{i} = (2000 kg)(11.11 \frac{m}{s}) = 22220kg.m/s \) northward. Final momentum, \( p_{f} \), when the truck started moving east is: \( p_{f} = (2000 kg)(13.89 \frac{m}{s}) = 27780kg.m/s \) eastward.

Calculate change in momentum

Since north and east are perpendicular directions the magnitudes of momenta can be added using Pythagorean theorem. The change in momentum: \( \Delta p = p_{f} - p_{i} \). They are at right angles so we get: \( \Delta p = \sqrt{{p_{f}}^{2} + {p_{i}}^{2}} = \sqrt{(27780)^{2} + (22220)^{2}}kg.m/s = 35507.22 kg.m/s\).

Find the direction of momentum change

Since we are dealing with vectors, we must also find the direction of the change in momentum. This means that we must find the angle the vector difference makes with the eastward direction. This is done by using trigonometry and solving for the angle \( \Theta \). Since Tangent of \( \Theta \) is equal to the ratio of the northwards (initial) momentum to the eastwards (final) momentum, we can write: \( Tan(\Theta) = \frac{p_{i}}{p_{f}} \). Solving for \( \Theta \) we find: \( \Theta = atan \left(\frac{22220}{27780}\right) =38.66^{\circ}\).

Represent the Result

The result is then the magnitude of the momentum change and the direction from the east towards the north. This is represented as: 35507.22 kg.m/s northward from east 38.66 degrees.

Key concepts

Velocity Conversion

When we talk about velocity in physics, it's crucial to ensure that our measurements are in standard units. For physics calculations, especially in SI units, converting kilometers per hour (km/h) to meters per second (m/s) is often necessary.
To convert, you multiply the speed in km/h by \( \frac{1000}{3600} \). This factor accounts for:

• 1000 meters in a kilometer
• 3600 seconds in an hour

Let's look at an example:

• A velocity of 40 km/h becomes \( 40 \times \frac{1000}{3600} = 11.11 \) m/s.
• Similarly, 50 km/h converts to \( 50 \times \frac{1000}{3600} = 13.89 \) m/s.

Using standard units simplifies calculations, reduces errors, and ensures the results are universally understood. This small step makes a big difference in scientific work.

Momentum Calculation

Momentum is a key concept in physics, representing the quantity of motion an object possesses. It's calculated by multiplying the mass of an object by its velocity. The formula for momentum \( p \) is:\[ p = mv \]where \( m \) is the mass in kilograms and \( v \) is the velocity in meters per second.
For example, if a truck has a mass of 2000 kg and a velocity of 11.11 m/s, its momentum is:\[ p = 2000 \times 11.11 = 22220 \] kg.m/s.
In our problem, the truck shifts directions, changing its velocity and thereby altering its momentum. Such calculations help us understand how forces act on moving objects and how these changes affect their path.
The change in momentum gives us insight into how much force was applied to change the truck's motion.

Vector Addition

Vectors have both magnitude and direction, making their addition different from regular numbers. When adding vectors, like initial and final momentum, you consider both these aspects.
In our case, the truck changes direction from north to east, making the momenta perpendicular vectors which can be added using the Pythagorean theorem.

• If the northward momentum is \( p_i = 22220 \) kg.m/s,
• and the eastward momentum is \( p_f = 27780 \) kg.m/s,

The magnitude of the change in momentum is:\[ \Delta p = \sqrt{(p_f)^2 + (p_i)^2} = \sqrt{(27780)^2 + (22220)^2} \approx 35507.22 \] kg.m/s.
This method provides a clearer picture of vector magnitude when each component is perpendicular, simplifying what could otherwise be a complex calculation.

Trigonometry in Physics

Trigonometry helps us find direction when dealing with vector quantities. When vectors have a directional change, like the truck's momentum change from north to east, trigonometric functions come into play.
To find the angle \( \Theta \) of the resultant vector, we use the tangent function:\[ \tan(\Theta) = \frac{p_i}{p_f} = \frac{22220}{27780} \]
By solving for \( \Theta \), we find:\[ \Theta = \arctan\left(\frac{22220}{27780}\right) \approx 38.66^\circ \]
This angle tells us how much the direction has shifted from east towards north. Understanding these directional changes is crucial in physics, providing not just the value but the orientation too, which is vital to predicting the future path of objects.