Question

Bats are extremely adept at catching insects in midair. If a 48.5-g bat flying in one direction at 8.51 m/s catches an 8.43-g insect flying in the opposite direction at 6.51 m/s, what is the speed of the bat immediately after catching the insect?

Answer 1

The speed of the bat immediately after catching the insect, found by applying the principle of conservation of momentum, is approximately 6.28 m/s.

Step-by-step explanation:

The student's question requires an application of conservation of momentum in physics. To find the speed of the bat immediately after catching the insect, we assume a perfectly inelastic collision where the bat and insect stick together. The total momentum before the collision must equal the total momentum after the collision.

Momentum before = Momentum after
(Mass of bat * Velocity of bat) + (Mass of insect * Velocity of insect) = (Total mass) * Velocity after

Let's plug in the values:

(0.0485 kg * 8.51 m/s) + (0.00843 kg * -6.51 m/s) = (0.0485 kg + 0.00843 kg) * Velocity after

Velocity after = [(0.0485 kg * 8.51 m/s) + (0.00843 kg * -6.51 m/s)] / (0.0485 kg + 0.00843 kg)

Velocity after = (0.412485 kg∗m/s - 0.0548873 kg∗m/s) / 0.05693 kg

Velocity after ≈ 6.28 m/s,

Therefore, the speed of the bat immediately after catching the insect is approximately 6.28 m/s.