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Chapter 7: Problem 58

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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