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A \(53.53-\mathrm{kg}\) boy is playing rollerblade dodgeball. His rollerblades are frictionless, and he is initially at rest. A dodgeball with a mass of \(513.1 \mathrm{~g}\) is thrown directly at him, and he catches it. After catching the ball, the boy moves with a speed of \(0.2304 \mathrm{~m} / \mathrm{s}\). With what speed was the dodgeball thrown?

Short Answer

Expert verified
Based on the given information and the conservation of momentum law, the initial velocity of the dodgeball when it was thrown is approximately 24.22 m/s.

Step by step solution

01

Convert the dodgeball mass to kg

We have the mass of the dodgeball as 513.1 g. To work with SI units, we need to convert it to kg. We can do this by dividing its mass by 1000: Dodgeball mass: \(\frac{513.1}{1000} = 0.5131 \mathrm{~kg}\)
02

Write down the conservation of momentum equation

According to the conservation of momentum law, Initial total momentum = Final total momentum \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f\) Here, \(m_1\) is the mass of the boy, \(v_1\) is the initial velocity of the boy, \(m_2\) is the mass of the dodgeball, \(v_2\) is the initial velocity of the dodgeball, and \(v_f\) is the final velocity of both after the collision. Since the boy is initially at rest, \(v_1 = 0\). We are given the final velocity \(v_f = 0.2304 \mathrm{~m/s}\) and we found out the mass of dodgeball in step 1. We can rearrange the equation to solve for the initial velocity of the dodgeball, \(v_2\).
03

Rearrange the equation to solve for the initial velocity of the dodgeball

The conservation of momentum equation can be rearranged to solve for the initial velocity of the dodgeball, \(v_2\): \(v_2 = \frac{(m_1 + m_2) v_f - m_1 v_1}{m_2}\) Now, we plug in the given values and solve for \(v_2\).
04

Calculate the initial velocity of the dodgeball

We substitute the given values and calculate \(v_2\): \(v_2 = \frac{(53.53\mathrm{~kg} + 0.5131\mathrm{~kg})(0.2304\mathrm{~m/s}) - 53.53\mathrm{~kg}(0)}{0.5131\mathrm{~kg}}\) \(v_2 = \frac{12.4257\mathrm{~kg\cdot m/s}}{0.5131\mathrm{~kg}}\) \(v_2 = 24.222\mathrm{~m/s}\) So, the initial velocity of the dodgeball when it was thrown is approximately \(24.22 \mathrm{~m/s}\).

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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 fundamental concept in physics that combines an object's mass and its velocity to represent its motion. The momentum of an object is calculated by the formula \( p = m \times v \), where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity. In the context of our dodgeball scenario, the principle of conservation of momentum is key. Before and after the boy catches the dodgeball, the total momentum must remain the same, provided no external forces act upon the system.

This principle allows us to set up the equation \( m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \) as seen in Step 2 of the solution. The next step is to input the right values into this equation, taking care to convert any measurements to SI units to ensure consistency. For instance, when we talk about velocity, we want to make sure we're using meters per second (m/s), and for mass, the kilograms (kg) unit is standard. Only by using the correct units can we ensure the accuracy of our momentum calculation.
Initial Velocity
Initial velocity is the speed at which an object is moving when it first comes under consideration. In physics problems, determining this value often involves understanding the conditions at the start of a scenario, before any changes in motion occur. In the dodgeball example, the key requirement is to find the speed at which the dodgeball was thrown, known as its initial velocity \( v_2 \).

In scenarios where the conservation of momentum applies, we can find the initial velocity by rearranging our momentum conservation equation, putting \( v_2 \) on one side. Given that the boy was initially at rest (\( v_1 = 0 \)), it reduces the complexity of our calculation. Following this, we substitute known values for the boy's mass (\( m_1 \)), the combined final velocity (\( v_f \)), and the mass of the dodgeball (\( m_2 \) - after converting it to kilograms) to find the initial velocity. The execution of this step was beautifully illustrated in Step 3 and Step 4 of the solution.
Mass Conversion
Mass conversion is a simple yet important aspect of solving physics problems, as it ensures that calculations are done using the consistent and correct units, typically kilograms in the International System of Units (SI). In the dodgeball problem, we encounter the need for mass conversion at the very start. The mass of the dodgeball is initially given in grams, which is not the SI unit for mass.

As such, a crucial first step was converting the dodgeball's mass from grams to kilograms by dividing by 1000, as seen in Step 1 of the solution (\( \frac{513.1}{1000} = 0.5131 \mathrm{~kg} \)). This conversion ensures compatibility when we apply the conservation of momentum equation. Without such conversions, our calculations could lead to incorrect results, potentially confusing the difference between mass and weight as well. Always remember to convert all mass measurements to kilograms when preparing to perform any momentum-related calculations in physics.

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Most popular questions from this chapter

A skateboarder of mass \(35.0 \mathrm{~kg}\) is riding her \(3.50-\mathrm{kg}\) skateboard at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). She jumps backward off her skateboard, sending the skateboard forward at a speed of \(8.50 \mathrm{~m} / \mathrm{s}\). At what speed is the skateboarder moving when her feet hit the ground?

Current measurements and cosmological theories suggest that only about \(4.6 \%\) of the total mass of the universe is composed of ordinary matter. About 23.96 of the mass is composed of dark matter, which does not emit or reflect light and can only be observed through its gravitational interaction with its surroundings (see Chapter 12). Suppose a galaxy with mass \(M_{\mathrm{G}}\) is moving in a straight line in the \(x\) -direction. After it interacts with an invisible clump of dark matter with mass \(M_{\mathrm{D} \mathrm{y}}\), the galaxy moves with 50.96 of its initial speed in a straight line in a direction that is rotated by an angle \(\theta\) from its initial velocity. Assume that initial and final velocities are given for positions where the galaxy is very far from the clump of dark matter, that the gravitational attraction can be neglected at those positions, and that the dark matter is initially at rest. Determine \(M_{\text {py }}\) in terms of \(M_{C}, v_{0}\) and \(\theta\).

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A 170.-g hockey puck moving in the positive \(x\) -direction at \(30.0 \mathrm{~m} / \mathrm{s}\) is struck by a stick at time \(t=2.00 \mathrm{~s}\) and moves in the opposite direction at \(25.0 \mathrm{~m} / \mathrm{s}\). If the puck is in contact with the stick for \(0.200 \mathrm{~s}\), plot the momentum and the position of the puck, and the force acting on it as a function of time, from 0 to \(5.00 \mathrm{~s}\). Be sure to label the coordinate axes with reasonable numbers.

Consider two carts, of masses \(m\) and \(2 m,\) at rest on a frictionless air track. If you push the lower-mass cart for \(35 \mathrm{~cm}\) and then the other cart for the same distance and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change. b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.

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