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Two physics students are doing a side competition during a game of bowling, seeing who can toss a ball with the larger momentum. The first bowler throws a \(4.5 \mathrm{kg}\) ball at \(7.2 \mathrm{m} / \mathrm{s}\). A second bowler throws a \(6.4 \mathrm{kg}\) ball. What speed must she beat to win the competition?

Short Answer

Expert verified
The second bowler must beat a speed of approximately 5.06 m/s in order to win the competition.

Step by step solution

01

Determine the Momentum of the First Bowler

Using the formula for momentum, the momentum of the first bowler's ball can be determined by multiplying the ball's mass by its velocity. So the momentum, \(M1\) of the first ball is calculated as \(M1=4.5kg \times 7.2m/s = 32.4kgm/s\).
02

Determine the Speed for the Second Bowler to Beat

In order for the second bowler to win the competition, she must throw her ball with a momentum larger than 32.4 kgm/s. We can use the formula for momentum, re-arrange it to solve for velocity. Using the mass of the second bowler's ball \(6.4kg\), we get \(velocity = momentum/mass = M1/6.4kg = 32.4kgm/s / 6.4kg\)
03

Calculate the Desired Speed

Carrying out the division gives approximately \(5.06m/s\). So, the second bowler needs to achieve a speed greater than 5.06 m/s to win the competition.

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Key Concepts

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

Physics Momentum Problem
When students encounter a physics momentum problem, such as comparing the momentum of two bowling balls thrown by different bowlers, a fundamental physics concept comes into play. Momentum, denoted typically as 'p,' is a measure of the motion of an object and is calculated as the product of an object's mass and its velocity. In mathematical terms, the formula is expressed as
\[ p = mv \]
where \(p\) represents momentum, \(m\) is the mass of the object, and \(v\) is its velocity.

Understanding that momentum is a vector quantity, which means it has both magnitude and direction, is crucial. In the bowling scenario, the direction is usually not factored because both balls are moving along the same path, allowing us to focus on the magnitude of momentum for comparison.

In problems like this, breaking down the exercise into manageable steps is essential. For students to solve the problem efficiently, it helps to start by calculating the momentum of the first object (or bowler's ball) using the given mass and velocity, as seen in the textbook solution's Step 1.
Conservation of Momentum
The conservation of momentum is a profound principle stating that within a closed system, the total momentum remains constant, provided no external forces act upon the objects involved. This is a cornerstone in physics and helps explain countless phenomena in classical mechanics.

The Law of Conservation of Momentum

Within the closed system, when two objects interact, such as colliding or separating molecules in a gas, the sum of their momenta before the interaction is equal to the sum of their momenta after the interaction. In a mathematical formula, it is represented as:
\[ p_{initial} = p_{final} \]
Here, \(p_{initial}\) signifies the total momentum of all objects before the interaction, while \(p_{final}\) represents the total momentum after.

This principle doesn't apply directly to the bowling ball problem since we're not dealing with an interaction between two objects. However, understanding momentum conservation is a building block for students in physics, providing a foundation for tackling more complex problems where multiple objects interact.
Calculating Velocity in Momentum
In the context of calculating velocity in momentum problems, once the momentum of an object is known, rearranging the momentum formula allows us to solve for velocity when the mass is also known.

Understanding the Momentum-Velocity Relationship

As seen in the initial steps of the solution, the formula \(p = mv\) can be re-written to isolate velocity (\(v\)) like so:
\[ v = \frac{p}{m} \]
In this format, by dividing the momentum by the mass, you obtain the velocity.

This relationship is what allows us to determine how fast the second bowler must throw her ball. Given a certain amount of momentum to beat and knowing the mass of her ball, the needed velocity is calculated, as seen in Steps 2 and 3 of the textbook solution. To improve your learning on such problems, focus on the fact that for a heavier object to have the same momentum as a lighter one, it must move at a slower speed, as momentum is directly dependent on both mass and velocity.

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Most popular questions from this chapter

III \(\mathrm{A} 2.7 \mathrm{kg}\) block of wood sits on a frictionless table. \(\mathrm{A} 3.0 \mathrm{g}\) bullet, fired horizontally at a speed of \(500 \mathrm{m} / \mathrm{s},\) goes completely through the block, emerging at a speed of \(220 \mathrm{m} / \mathrm{s}\). What is the speed of the block immediately after the bullet exits?

Casey is driving a \(1600 \mathrm{kg}\) car toward the east. She goes through an intersection at a speed of \(16 \mathrm{m} / \mathrm{s}\) (approximately \(35 \mathrm{mph}),\) the speed limit on both roads of the intersection. Kerry is driving a car of mass \(1200 \mathrm{kg}\) into the intersection, going north, and doesn't see or doesn't heed a red light, and slams into Casey's car. The cars lock together and skid to a stop. Later, the two review the scene with the police. Skid marks from the instant after the collision reveal that the two cars were moving exactly northeast. Kerry claims to have been driving at the speed limit, but Casey says that Kerry seemed to be going over the speed limit before the collision. Who is correct? Use the concept of conservation of momentum to make your case.

A student throws a 120 g snowball at \(7.5 \mathrm{m} / \mathrm{s}\) at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is \(0.15 \mathrm{s} ?\)

A water pipe in a building delivers 1000 liters (with mass \(1000 \mathrm{kg}\) ) of water per second. The water is moving through the pipe at \(1.4 \mathrm{m} / \mathrm{s}\). The pipe has a \(90^{\circ}\) bend, and the pipe will require a supporting structure, called a thrust block, at the bend, as in Figure \(\mathrm{P} 9.70\). We can use the ideas of momentum and impulse to understand why. Each second, \(1000 \mathrm{kg}\) of water moving at \(v_{x}=1.4 \mathrm{m} / \mathrm{s}\) changes direction to move at \(v_{y}=1.4 \mathrm{m} / \mathrm{s}\) a. What are the magnitude and direction of the change in momentum of the \(1000 \mathrm{kg}\) of water? b. What are the magnitude and direction of the necessary impulse? c. This impulse takes place over 1.0 s. What is the necessary force?

Ferns spread spores instead of seeds, and some ferns eject the spores at surprisingly high speeds. One species accelerates \(1.4 \mu \mathrm{g}\) spores to a \(4.5 \mathrm{m} / \mathrm{s}\) ejection speed in a time of \(1.0 \mathrm{ms}\) What impulse is provided to the spores? What is the average force on a spore?

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