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Bats are extremely adept at catching insects in midair. If a 50.0-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: The speed of the bat immediately after catching the insect is approximately 6.73 m/s.

Step by step solution

01

Identify known and unknown quantities

We are given the following information: - Mass of the bat (m1) = 50.0 g = 0.050 kg (converted to kilograms) - Initial speed of the bat (v1) = 8.00 m/s - Mass of the insect (m2) = 5.00 g = 0.005 kg (converted to kilograms) - Initial speed of the insect (v2) = -6.00 m/s (negative sign indicates opposite direction) We need to find the final speed (v') of the bat-insect system after the collision.
02

Write the conservation of linear momentum equation

Since there is no external force acting on the system, the total momentum before the collision is equal to the total momentum after the collision. The equation for conservation of linear momentum is: m1 * v1 + m2 * v2 = (m1 + m2) * v'
03

Plug in the known values in the equation

Now we can plug in the given values into the equation: (0.050 kg) * (8.00 m/s) + (0.005 kg) * (-6.00 m/s) = (0.050 kg + 0.005 kg) * v'
04

Solve for the final speed (v') of the bat-insect system

First, calculate the left-hand side of the equation: (0.050 kg) * (8.00 m/s) + (0.005 kg) * (-6.00 m/s) = 0.400 kg m/s - 0.030 kg m/s = 0.370 kg m/s Next, solve for v': 0.370 kg m/s = (0.055 kg) * v' v' = (0.370 kg m/s) / (0.055 kg) v' ≈ 6.73 m/s
05

Write the final answer

The speed of the bat immediately after catching the insect is approximately 6.73 m/s.

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Key Concepts

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

Collision
In the world of physics, a collision refers to an event where two or more bodies exert forces on each other in a relatively short period. Even though we associate collisions with dramatic crashes, they aren't always destructive.
Instead, they can involve gentle interactions, like a bat catching an insect mid-flight. In our exercise, the collision between a bat and an insect is an example of an inelastic collision because the bat and the insect move together as one system after the interaction.
The concept of a collision is crucial because it involves interactions where momentum might be transferred or shared between objects. Such interactions help us understand concepts like energy conservation and momentum transfer.
A key thing to remember is that not all collisions involve sticking together. Some are elastic, where objects bounce off each other, conserving both momentum and kinetic energy.
Linear Momentum
Linear momentum is a fundamental concept in physics that describes the motion of an object. It is a measure of an object's mass in motion. The formula to calculate linear momentum is given by: \[ p = m imes v \]where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity.In terms of our exercise, before the bat catches the insect, both have their momentum.
  • The bat's momentum: \( 0.050 \, \text{kg} \times 8.00 \, \text{m/s} = 0.400 \text{ kg m/s} \)
  • The insect's momentum: \( 0.005 \, \text{kg} \times -6.00 \, \text{m/s} = -0.030 \text{ kg m/s} \)
The law of conservation of momentum, which states that the total momentum of an isolated system remains constant if no external forces are acting, is applicable in this problem. This principle allows us to calculate the final velocity of the combined bat-insect system after the collision. It beautifully illustrates how momentum is shared and conserved in an interaction.
Velocity
Velocity is a vector quantity, which means it has both magnitude and direction. It is fundamentally different from speed, which only considers magnitude. In our exercise, understanding the velocity of each object, both before and after the collision, is vital to solving for the final answer.
Initially:
  • The bat's velocity is \( 8.00 \, \text{m/s} \) toward the insect.
  • The insect's velocity is \( -6.00 \, \text{m/s} \), indicating it's moving in the opposite direction.
Upon collision, the resulting velocity of the bat and insect system can be derived using conservation principles. The negative sign for the insect's velocity shows the opposite direction relative to the bat, making it necessary to consider the direction when dealing with vector quantities. This true understanding of velocity helps identify the resulting direction and speed after the collision. **So, the final velocity of this system is calculated as approximately \( v' \approx 6.73 \, \text{m/s} \), indicating that the bat continues moving forward after catching the insect.**

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Most popular questions from this chapter

Stuck in the middle of a frozen pond with only your physics book, you decide to put physics in action and throw the 5.00 -kg book. If your mass is \(62.0 \mathrm{~kg}\) and you throw the book at \(13.0 \mathrm{~m} / \mathrm{s}\), how fast do you then slide across the ice? (Assume the absence of friction.)

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.

In waterskiing, a "garage sale" occurs when a skier loses control and falls and waterskis fly in different directions. In one particular incident, a novice skier was skimming across the surface of the water at \(22.0 \mathrm{~m} / \mathrm{s}\) when he lost control. One ski, with a mass of \(1.50 \mathrm{~kg},\) flew off at an angle of \(12.0^{\circ}\) to the left of the initial direction of the skier with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). The other identical ski flew from the crash at an angle of \(5.00^{\circ}\) to the right with a speed of \(21.0 \mathrm{~m} / \mathrm{s} .\) What was the velocity of the \(61.0-\mathrm{kg}\) skier? Give a speed and a direction relative to the initial velocity vector.

An uncovered hopper car from a freight train rolls without friction or air resistance along a level track at a constant speed of \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. The mass of the car is \(1.18 \cdot 10^{5} \mathrm{~kg}\). a) As the car rolls, a monsoon rainstorm begins, and the car begins to collect water in its hopper (see the figure). What is the speed of the car after \(1.62 \cdot 10^{4} \mathrm{~kg}\) of water collects in the car's hopper? Assume that the rain is falling vertically in the negative \(y\) -direction. b) The rain stops, and a valve at the bottom of the hopper is opened to release the water. The speed of the car when the valve is opened is again \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction (see the figure). The water drains out vertically in the negative \(y\) -direction. What is the speed of the car after all the water has drained out?

After several large firecrackers have been inserted into its holes, a bowling ball is projected into the air using a homemade launcher and explodes in midair. During the launch, the 7.00 -kg ball is shot into the air with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\) at a \(40.0^{\circ}\) angle; it explodes at the peak of its trajectory, breaking into three pieces of equal mass. One piece travels straight up with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Another piece travels straight back with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). What is the velocity of the third piece (speed and direction)?

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