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Chapter 9: Problem 19

A \(2100 \mathrm{~kg}\) truck traveling north at \(41 \mathrm{~km} / \mathrm{h}\) turns east and accelerates to \(51 \mathrm{~km} / \mathrm{h}\). (a) What is the change in the truck's kinetic energy? What are the (b) magnitude and (c) direction of the change in its momentum?

Short Answer

Expert verified
(a) 74,216.4 J; (b) 37,983 kg m/s; (c) 51.8掳 east of north.

Step by step solution

01

Convert Speeds to m/s

First, we'll convert the speeds from km/h to m/s. The initial speed is 41 km/h and the final speed is 51 km/h. We use the conversion factor: \[ \text{Speed (m/s)} = \text{Speed (km/h)} \times \frac{1000}{3600} \]Initial speed:\[ 41 \times \frac{1000}{3600} = 11.39 \text{ m/s} \]Final speed:\[ 51 \times \frac{1000}{3600} = 14.17 \text{ m/s} \]
02

Calculate Initial and Final Kinetic Energy

Kinetic energy is given by the formula:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass and \( v \) is the velocity.Initial kinetic energy:\[ KE_i = \frac{1}{2} \times 2100 \times (11.39)^2 = 136,192.05 \text{ J} \]Final kinetic energy:\[ KE_f = \frac{1}{2} \times 2100 \times (14.17)^2 = 210,408.45 \text{ J} \]
03

Calculate Change in Kinetic Energy

The change in kinetic energy, \( \Delta KE \), is the difference between the final and initial kinetic energies:\[ \Delta KE = KE_f - KE_i = 210,408.45 - 136,192.05 = 74,216.4 \text{ J} \]
04

Calculate Initial and Final Momentum

Momentum is given by the formula:\[ p = mv \]Initial momentum (northward):\[ p_i = 2100 \times 11.39 = 23,919 \text{ kg m/s} \]Final momentum (eastward):\[ p_f = 2100 \times 14.17 = 29,757 \text{ kg m/s} \]
05

Calculate Magnitude of Change in Momentum

The change in momentum involves a vector change, so we use the Pythagorean theorem:\[ \Delta p = \sqrt{(p_f)^2 + (p_i)^2} = \sqrt{(29,757)^2 + (23,919)^2} = 37,983 \text{ kg m/s} \]
06

Determine Direction of Change in Momentum

We calculate the angle using tangent:\[ \tan(\theta) = \frac{p_f}{p_i} = \frac{29,757}{23,919} \]\[ \theta = \tan^{-1}\left(\frac{29,757}{23,919}\right) \approx 51.8^\circ \]Thus, the change in momentum is directed to the northeast at approximately 51.8 degrees east of north.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum
Momentum is a key concept in physics, especially when it comes to moving objects like vehicles. It is a measure of the motion an object has and is calculated as the product of an object's mass and velocity. The formula used is:
\[ p = mv \]
In this equation:
  • \( p \) stands for momentum
  • \( m \) is the mass of the object (in kg)
  • \( v \) is its velocity (in m/s)
Momentum describes how much force is needed to stop an object. An object with a large mass or high velocity will have significant momentum. In our exercise, we calculate the truck's initial momentum when it is moving northward and final momentum when it turns and accelerates eastward. Understanding changes in momentum helps in analyzing different motion scenarios.
Pythagorean theorem
The Pythagorean theorem is a fundamental principle used in geometry. It relates the lengths of the sides of right-angled triangles. The theorem states that for a right triangle:
\[ a^2 + b^2 = c^2 \]
Where:
  • \( a \) and \( b \) are the lengths of the triangle's two shorter sides
  • \( c \) is the length of the hypotenuse, the side opposite the right angle
In the context of our problem, when the truck changes its direction from north to east, the components of momentum form a right triangle. The magnitude of the momentum change is calculated using this theorem, treating the north and east directions as perpendicular components. By plugging the values into the Pythagorean theorem, we find the total change in vector magnitude.
Vector change
When an object changes direction, its velocity and consequently its momentum change occurs in a vector form, not just in magnitude but in direction too. A vector has both magnitude and direction, which makes calculations slightly more complex than simply adding numbers.
Consider the truck's case which changes its motion from north to east:
  • Its initial momentum vector is northward, and final momentum is eastward.
  • These two movements create a vector change, which can be represented by constructing a right triangle using initial and final vectors.
  • The vector change is the hypotenuse found using the Pythagorean theorem, showing the total momentum change in both direction and magnitude.
Understanding vector changes is crucial in physics for calculating resultant forces, movements, and predicting outcomes.
Angle calculation
Direction in physics is often described using angles, especially when discussing vector movements. To find the angle of the changed momentum direction, we employ trigonometric functions like tangent (tan), which is particularly helpful when you have opposite and adjacent sides in a right triangle.
The formula is:
\[ \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \]
In our scenario:
  • The opposite side is the final momentum eastward.
  • The adjacent side is the initial momentum northward.
By calculating \( \tan^{-1} \) of their ratio, we find the angle \( \theta \), which in this case is approximately 51.8 degrees. This angle helps describe the direction of the vector change, providing insight into how far the truck's direction changed.

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Most popular questions from this chapter

block 1 (mass \(2,0 \mathrm{~kg}\) ) is moving rightward at \(10 \mathrm{~m} / \mathrm{s}\) and block 2 (mass \(5.0 \mathrm{~kg}\) ) is moving rightward at \(3.0 \mathrm{~m} / \mathrm{s}\) The surface is frictionless, and a spring with a spring constant of \(1120 \mathrm{~N} / \mathrm{m}\) is fixed to block 2 . When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

A \(20.0 \mathrm{~kg}\) body is moving through space in the positive dircction of an \(x\) axis with a speed of \(200 \mathrm{~m} / \mathrm{s}\) when, due to an internal explosion, it breaks into three parts. One part, with a mass of \(10.0 \mathrm{~kg}\), moves away from the point of explosion with a speed of \(100 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. A second part, with a mass of \(4.00 \mathrm{~kg}\), moves in the negative \(x\) direction with a speed of \(500 \mathrm{~m} / \mathrm{s}\). (a) In unit-vector notation, what is the velocity of the third part? (b) How much cnergy is released in the explosion? Ignore effects due to the gravitational force.

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with the cue ball's original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

A uniform soda can of mass \(0.140 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) tall and filled with \(0.354 \mathrm{~kg}\) of soda (Fig. \(9-41\) ). Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height \(h\) of the com of the can and contents (a) initially and (b) after the can loses all the soda? (c) What happens to \(h\) as the soda drains out? (d) If \(x\) is the height of the remaining soda at any given instant, find \(x\) when the com rcaches its lowest point.
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