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Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter (d) the Earth (mass running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The momentum for (a) the proton is approximately \(8.35 \times 10^{-21}\) kg m/s; (b) the bullet is \(4.5\) kg m/s; (c) the sprinter is \(750\) kg m/s; and (d) the Earth is approximately \(1.78 \times 10^{29}\) kg m/s.

Step by step solution

01

Calculate Proton Momentum

The momentum of the proton can be calculated using the formula \(p=mv\), where \(m = 1.67 \times 10^{-27}\) kg is the mass and \(v = 5.00 \times 10^{6} \) m/s is the velocity. Evaluate this product to find the momentum.
02

Calculate Bullet Momentum

The momentum of the bullet is calculated in a similar manner. Use the given mass \(m = 15.0\) g (converted to kg by multiplying by 0.001 to get 0.015 kg) and velocity \(v = 300\) m/s in the same formula to find the momentum.
03

Calculate Sprinter Momentum

Again use the same formula to calculate the momentum for the sprinter. The given mass is \(m = 75.0\) kg and velocity is \(v = 10.0\) m/s.
04

Calculate Earth Momentum

For the Earth, the mass is given as \(m = 5.98 \times 10^{24}\) kg and the velocity as \(v = 2.98 \times 10^{4}\) m/s. Use these values in the formula to calculate the Earth's momentum.

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Key Concepts

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

Momentum Calculation
Momentum is a key physical quantity that measures the motion of an object. It is defined as the product of an object's mass and its velocity. The unit of momentum in the International System (SI) is kilogram meter per second (kg·m/s). The formula for calculating linear momentum is:
\[ p = m \times v \]
where:
  • \( p \) is the momentum,
  • \( m \) is the mass of the object, and
  • \( v \) is the velocity of the object.
This formula tells us that the momentum of an object increases as its mass or velocity increases. Understanding this helps us to solve problems relating to objects in motion, such as particles, vehicles, and celestial bodies. Now, let's explore specific examples of momentum.
Proton Momentum
In physics, protons are fundamental subatomic particles with a positive charge. They have a very small mass on the atomic scale. Therefore, calculating the momentum of a proton involves small numbers. For example, when a proton with a mass of \(1.67 \times 10^{-27}\) kg moves at a speed of \(5.00 \times 10^{6}\) m/s, its momentum \( p \) can be calculated as:
\[ p = (1.67 \times 10^{-27} \text{ kg}) \times (5.00 \times 10^{6} \text{ m/s}) \]
Evaluating the product, we obtain a momentum of \(8.35 \times 10^{-21}\) kg·m/s. This small value reflects the tiny mass of the proton. It's important for understanding behaviors at the atomic scale and plays a critical role in areas like particle physics.
Bullet Momentum
Bullets, with their relatively small mass, achieve high values of momentum due to their high speeds. To compute the momentum of a bullet weighing 15.0 grams moving at 300 m/s, we need to first convert the mass to kilograms, yielding 0.015 kg. Using the same momentum formula, we compute:
\[ p = (0.015 \text{ kg}) \times (300 \text{ m/s}) \]
The resulting momentum is \(4.5\) kg·m/s. This illustrates how even small objects can have significant momentum due to their velocity. In practical terms, understanding bullet momentum is crucial in fields such as ballistics, which explore the dynamics of projectiles.
Earth Momentum
The Earth, despite moving relatively slowly compared to subatomic particles, possesses immense momentum due to its enormous mass. With a massive weight of \(5.98 \times 10^{24}\) kg and moving through space at an impressive speed of \(2.98 \times 10^{4}\) m/s, Earth's momentum can be calculated as:
\[ p = (5.98 \times 10^{24} \text{ kg}) \times (2.98 \times 10^{4} \text{ m/s}) \]
This calculation results in a colossal momentum of about \(1.78 \times 10^{29}\) kg·m/s. This massive momentum contributes significantly to the gravitational interactions and dynamics observed in our solar system, impacting everything from orbital stability to the effects of gravitational pulls.

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Most popular questions from this chapter

A \(0.10\) -kg ball is thrown straight up into the air with an initial speed of \(15 \mathrm{~m} / \mathrm{s}\). Find the momentum of the ball (a) at its maximum height and (b) halfway to its maximum height.

ecp Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted onthe truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{~m} / \mathrm{s}\) and that they undergo a perfectly inelastic headon collision. Each driver has mass \(80.0 \mathrm{~kg} .\) Including the masses of the drivers, the total masses of the vehicles are \(800 \mathrm{~kg}\) for the car and \(4000 \mathrm{~kg}\) for the truck. If the collision time is \(0.120 \mathrm{~s}\), what force does the seat belt exert on each driver?

A ball of mass \(0.150 \mathrm{~kg}\) is dropped from rest from a height of \(1.25 \mathrm{~m}\). It rebounds from the floor to reach a height of \(0.960 \mathrm{~m}\). What impulse was given to the ball by the floor?

A \(12.0-\mathrm{g}\) bullet is fired horizontally into a \(100-\mathrm{g}\) wooden block initially at rest on a horizontal surface. After impact, the block slides \(7.5 \mathrm{~m}\) before coming to rest. If the coefficient of kinetic friction between block and surface is \(0.650\), what was the speed of the bullet immediately before impact?

An unstable nucleus of mass \(1.7 \times 10^{-26} \mathrm{~kg}\), initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of \(m_{1}=5.0 \times 10^{-27} \mathrm{~kg}\), moves in the positive \(y\) -direction with speed \(v_{1}=6.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Another particle, of mass \(m_{2}=8.4 \times 10^{-27} \mathrm{~kg}\), moves in the positive \(x\) -direction with speed \(v_{2}=4.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the velocity of the third particle.

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