Problem 58 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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'

Step 3: 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'

Step 4: 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

Step 5: Write the final answer

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

Key Concepts

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.**