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Bats are extremely adept at catching insects in midair. If a \(50.0-\mathrm{g}\) bat flying in one direction at \(8.00 \mathrm{~m} / \mathrm{s}\) catches a \(5.00-\mathrm{g}\) insect flying in the opposite direction at \(6.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the bat immediately after catching the insect?

Short Answer

Expert verified
Answer: 6.727 m/s

Step by step solution

01

Write down the law of conservation momentum formula

The law of conservation of momentum states that the total initial momentum of the system should be equal to the total final momentum of the system. The formula can be written as: \(m_1v_1 + m_2v_2 = (m_1 + m_2)V_{f}\), where \(m_1\) and \(m_2\) are the masses of the bat and insect respectively, \(v_1\) and \(v_2\) are their initial velocities, and \(V_{f}\) is the final velocity of the system (the speed of the bat immediately after catching the insect).
02

Plug in given values

Now, we can plug in the given mass and velocity values into the conservation of momentum formula: \((50.0\text{ g})(8.00\text{ m/s}) + (5.00\text{ g})(-6.00\text{ m/s}) = (50.0\text{ g} + 5.00\text{ g})V_{f}\). Remember that the insect's velocity is negative because it's flying in the opposite direction of the bat.
03

Solve for the final velocity \(V_{f}\)

Simplify the equation by carrying out the multiplication: \((400\text{ g m/s}) + (-30\text{ g m/s}) = (55.0\text{ g})V_{f}\). Now, solve for the final velocity \(V_{f}\): \(V_{f} = \frac{400\text{ g m/s} - 30\text{ g m/s}}{55.0\text{ g}} = \frac{370\text{ g m/s}}{55.0\text{ g}}\). Divide the numerator by the denominator: \(V_{f} = 6.727\text{ m/s}\), approximately. The speed of the bat immediately after catching the insect is approximately \(6.727\text{ m/s}\).

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Key Concepts

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

Momentum Formula
In physics, momentum refers to the measure of motion an object has and is a key concept in understanding how objects interact in collisions. The momentum formula is a mathematical representation of this concept and is given by the equation \( p = mv \), where \( p \) is the momentum, \( m \) is the mass of the object, and \( v \) is the velocity.

When two objects collide, as in our example where a bat catches an insect in midair, their momenta before and after the collision can tell us a great deal about the outcome of their interaction. The principle of conservation of momentum says that the total momentum of a closed system remains constant if no external forces are acting upon it. This principle lays the foundation for analyzing the motion of objects post-collision based on their masses and velocities prior to the collision.

Understanding this principle can significantly improve a student's ability to solve problems related to collision. It emphasizes the importance of recognizing when momentum is conserved and applying the formula correctly by including direction in the form of positive or negative values for velocity, depending on the object's direction of travel.
Mass and Velocity
The concepts of mass and velocity are fundamental to the study of momentum. Mass is a measure of how much 'stuff,' or matter, an object has and is typically measured in kilograms (kg) or grams (g). In contrast, velocity is a vector quantity that describes an object's speed and direction. It's measured in meters per second (m/s) or other units of speed.

Impact of Mass and Velocity

When considering momentum, one should note that an object’s final velocity after an interaction can be more significantly influenced by mass if the velocities are relatively small. In our example, the bat has a substantially greater mass than the insect, which implies that it can exert a greater influence on the final velocity of the system after they collide.

Additionally, the sign of velocity is crucial since it indicates direction. If objects are moving in opposite directions, one of the velocities must be negative in the calculation to reflect this. Students should be vigilant in assigning the correct direction to velocity values to achieve the correct solution for problems involving momentum.
Final Velocity Calculation
Calculating the final velocity of a system after a collision, as shown in the bat-insect problem, involves using the conservation of momentum formula. Once you know the masses and initial velocities of the objects involved, you use the formula \(m_1v_1 + m_2v_2 = (m_1 + m_2)V_f\), where \(V_f\) is the unknown final velocity you are solving for.

Steps to Calculate Final Velocity

To determine \(V_f\), you first need to ensure the mass units are consistent and sum the momenta of the individual objects before the collision. This is the left side of the equation. Then you divide by the combined mass of the two objects post-collision to find the final shared velocity. Remember to account for the correct direction of each initial velocity: if the objects are moving toward each other, their velocities will have opposite signs. A thoughtful step-by-step approach can help clarify the process:
  • Write down the conservation of momentum formula
  • Fill in the known values for mass and initial velocity
  • Perform the algebraic operations to isolate \(V_f\) on one side of the equation
  • Solve for \(V_f\) and interpret the result
Through practice and application of these concepts, students can cement their understanding of momentum and accurately calculate final velocities in a variety of scenarios.

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Most popular questions from this chapter

A Super Ball is dropped from a height of \(3.935 \mathrm{~m}\). Its maximum height on its third bounce is \(2.621 \mathrm{~m}\). What is the coefficient of restitution of the ball?

A student with a mass of 60.0 kg jumps straight up in the air by using her legs to apply an average force of \(770 .\) N to the ground for 0.250 s. Assume that the initial momentum of the student and the Earth are zero. What is the momentum of the student immediately after this impulse? What is the momentum of the Earth after this impulse? What is the speed of the Earth after the impulse? What fraction of the total kinetic energy that the student produces with her legs goes to the Earth (the mass of the Earth is \(5.98 \cdot 10^{24} \mathrm{~kg}\) )? Using conservation of energy, how high does the student jump?

Which of the following statements about car collisions is (are) true? a) The essential safety benefit of crumple zones (parts of the front of a car designed to receive maximum deformation during a head-on collision) is due to their absorbing kinetic energy, converting it into deformation, and lengthening the effective collision time, thus reducing the average force experienced by the driver. b) If car 1 has mass \(m\) and speed \(v,\) and car 2 has mass \(0.5 m\) and speed \(1.5 v\), then both cars have the same momentum. c) If two identical cars with identical speeds collide head on, the magnitude of the impulse received by each car and each driver is the same as if one car at the same speed had collided head on with a concrete wall. d) Car 1 has mass \(m,\) and car 2 has mass \(2 m .\) In a head-on collision of these cars while moving at identical speeds in opposite directions, car 1 experiences a bigger acceleration than car 2 . e) Car 1 has mass \(m,\) and car 2 has mass \(2 m .\) In a head-on collision of these cars while moving at identical speeds in opposite directions, car 1 receives an impulse of bigger magnitude than that received by car 2 .

In bocce, the object of the game is to get your balls (each with mass \(M=1.00 \mathrm{~kg}\) ) as close as possible to the small white ball (the pallina, mass \(m=0.0450 \mathrm{~kg}\) ). Your first throw positioned your ball \(2.00 \mathrm{~m}\) to the left of the pallina. If your next throw arrives with a speed of \(v=1.00 \mathrm{~m} / \mathrm{s}\) and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.200\), what are the final distances of your two balls from the pallina in each of the following cases? Assume that collisions are elastic. a) You throw your ball from the left, hitting your first ball. b) You throw your ball from the right, hitting the pallina.

A Super Ball has a coefficient of restitution of 0.8887 . If the ball is dropped from a height of \(3.853 \mathrm{~m}\) above the floor, what maximum height will it reach on its third bounce?

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