Bats are extremely adept at catching insects in midair. If a 62-g bat flying in one direction at 9 m/s catches a 6-g insect flying in the opposite direction at 4 m/s, what is the speed of the bat immediately after catching the insect?
Momentum = m1*V1-m2*V2 = 0.062*9-0.006*4 = 0.558 - 0.024 = 0.534
m1*V = 0.534
0.062*V = 0.534
V = 8.61 m/s.
To solve this problem, we can apply the principle of conservation of momentum. The momentum before the catch must be equal to the momentum after the catch.
The momentum of an object is given by the product of its mass and velocity. So, let's calculate the momentum before the catch for both the bat and the insect separately.
Momentum of the bat = mass of the bat × velocity of the bat
= 62 g × 9 m/s
= (62 × 10⁻³) kg × 9 m/s
= 0.558 kg·m/s
Momentum of the insect = mass of the insect × velocity of the insect
= 6 g × (-4) m/s (Negative sign indicates opposite direction)
= (6 × 10⁻³) kg × -4 m/s
= -0.024 kg·m/s
Now, since momentum is conserved, the total momentum before the catch should be equal to the total momentum after the catch.
Total momentum before = Total momentum after
Momentum of the bat + Momentum of the insect = Total momentum after
0.558 kg·m/s + (-0.024 kg·m/s) = Total momentum after
Simplifying this, we find,
Total momentum after = 0.534 kg·m/s
Now, to find the speed of the bat immediately after catching the insect, we need to divide the total momentum after catch by the mass of the bat.
Speed of the bat = Total momentum after / mass of the bat
Speed of the bat = 0.534 kg·m/s / 62 g
= (0.534 kg·m/s) / (62 × 10⁻³) kg
≈ 8.613 m/s
Therefore, the speed of the bat immediately after catching the insect is approximately 8.613 m/s.