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(a) What is the magnitude of the momentum of a \(10.000 \mathrm{~kg}\) truck whose speed is \(12.0 \mathrm{~m} / \mathrm{s} 7\) (b) What speed would a \(2000 \mathrm{~kg}\) SUV have to attain in order to have (i) the same momentum? (ii) the same kinctic energy?

Short Answer

Expert verified
The magnitude of the momentum of the truck is 120,000 kg·m/s. The SUV needs to attain a velocity of 60 m/s to have the same momentum as the truck, and a velocity of 30 m/s to have the same kinetic energy.

Step by step solution

01

Calculate the Momentum of the Truck

Momentum (p) is calculated by the formula \( p = m \cdot v \) where m is mass and v is velocity. Here, \( m = 10,000 \, kg \) and \( v = 12.0 \, m/s \), thus \( p_{truck} = 10,000 \, kg \cdot 12.0 \, m/s = 120,000 \, kg \cdot m/s \). Therefore, the magnitude of the momentum of the truck is 120,000 \( kg \cdot m/s \).
02

Determine the Velocity of the SUV to have the Same Momentum

In order to have the same momentum as the truck, the SUV's momentum (p_{SUV}) must also be 120,000 kg · m/s. Using the formula \( p = m \cdot v \), we can solve for the velocity of the SUV \((v_{suv}) = p_{SUV} / m_{suv} \), where \(m_{SUV} = 2000 \, kg\). Thus, \(v_{suv(momentum)} = 120,000 \, kg \cdot m/s \, / \, 2000 \, kg = 60 \, m/s \)
03

Determine the Velocity of the SUV to have the Same Kinetic Energy

Kinetic energy (KE) is calculated by the formula KE = 1/2 \cdot m \cdot v^{2}. The kinetic energy of the truck is KE_{truck} = 1/2 * 10,000 kg * (12 m/s)^{2} = 720,000 J. To find the velocity of the SUV to have the same kinetic energy, we use the formula to solve for velocity: \( v_{suv(KE)} = \sqrt{(2 \cdot KE_{suv} / m_{suv})} \). Substituting the values \( KE_{suv} = 720,000 \, J \) and \( m_{suv} = 2000 \, kg \), the velocity of the SUV to have the same kinetic energy as the truck is \( v_{suv(KE)} = \sqrt{(2 \cdot 720,000 \, J / 2000 \, kg)} = 30 \, m/s \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a type of energy that an object possesses due to its motion. It's one of the most fundamental ideas in physics. When any object moves, it has kinetic energy, which is calculated using the formula:\[ KE = \frac{1}{2} \cdot m \cdot v^2 \]- **m** represents the mass of the object,- **v** is the velocity of the object.This equation shows that kinetic energy increases quadratically with speed. That means if you double an object's speed, its kinetic energy increases four times. Amazing, right? In our example, both the truck and the SUV need to have the same kinetic energy. By comparing calculations, it was revealed that velocity plays a big role. An SUV, being lighter than the truck, must go faster to have the same kinetic energy.Understanding kinetic energy helps explain how energy is transferred and transformed from one body to another. This has vast applications, from daily life to advanced fields like engineering and space exploration.
Conservation of Momentum
Momentum, similar to kinetic energy, is a core concept of movement in physics. It's defined as the product of an object's mass and its velocity:\[ p = m \cdot v \]Momentum is a vector quantity, which means it has both magnitude and direction. The law of conservation of momentum states that the total momentum in an isolated system remains constant, provided no external forces are acting on it.- This is expressed as: - Before collision: Total momentum = After collision: Total momentum- For our calculation, the truck and SUV were considered separately. By calculating the velocity needed for the SUV to have the same momentum as the truck, we demonstrated that momentum can be conserved within any isolated system.This principle is foundational for understanding how objects interact with one another, especially in collisions. If you've ever played pool, you've seen this law in action. When a moving ball hits another, the momentum gets transferred. By understanding and applying conservation of momentum, we gain insight into a significant aspect of physical interactions.
Physics Calculations
Physics calculations are essential for solving problems about motion and energy. They allow us to predict, understand, and manipulate the natural world. These calculations often involve equations and formulas that let us quantify relationships between different physical quantities, such as mass, velocity, and energy.Let's delve into some key elements:- **Units and Dimensions:** Physics calculations need consistent units of measurement. For velocity, we use meters per second (\(m/s\)), for mass, kilograms (\(kg\)), and for energy, joules (\(J\)).- **Substitution:** Once the formula is identified, numbers are substituted in place of variables. In our exercise, we used substitutions to find the momentum and kinetic energy of the truck and SUV.- **Problem-solving:** Following a clear step-by-step approach is crucial. First, understand the problem, identify known and unknown variables, and then apply the appropriate formula.These calculations help harness complex principles into practical solutions. This structured approach not only fosters a deeper understanding of physics but also helps tackle real-life challenges in fields like engineering, architecture, and technology.

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Most popular questions from this chapter

A Variable-Mass Raindrop. In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$ F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t} $$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k u .\) This gives, since \(F_{\text {cut }}=m g_{*}\) $$ m g=m \frac{d v}{d t}+v(k v) $$ Or, dividing by \(k_{*}\) $$ x_{B}=x \frac{d v}{d t}+v^{2} $$ This is a differential equation that has a solution of the form \(v=a t\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v\). find the acccleration \(a\). (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{~s} .(\mathrm{c})\) Given that \(k=2.00 \mathrm{~g} / \mathrm{m},\) find the mass of the raindrop at \(t=3.00 \mathrm{~s}\). (For many more intriguing aspects of this problem, see K. S. Krane, American Jourmal of Physics, Vol. 49(1981) . pp. \(113-117 .\) )

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\(A 12.0 \mathrm{~kg}\) shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of \(150 \mathrm{~m} / \mathrm{s}\). At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was relcased in the explosion?

A baseball has mass \(0.145 \mathrm{~kg}\). (a) If the velocity of a pitched ball has a magnitude of \(45.0 \mathrm{~m} / \mathrm{s}\) and the batted ball's velocity is \(55.0 \mathrm{~m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for \(2.00 \mathrm{~ms}\), find the magnitude of the average force applied by the bat.

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