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Chapter 6: Problem 2

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

Expert verified
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

01

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 \( v2 = (50 \frac{km}{h})(\frac{1000 m}{1 km})(\frac{1 h}{3600 s}) = 13.89 \frac{m}{s} \).
02

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.
03

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: \( \Delta p = \sqrt{{p_{f}}^{2} + {p_{i}}^{2}} = \sqrt{(27780)^{2} + (22220)^{2}}kg.m/s = 35507.22 kg.m/s\).
04

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: \( \Theta = atan \left(\frac{22220}{27780}\right) =38.66^{\circ}\).
05

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.

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Most popular questions from this chapter

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