Question

The hog-nosed bat is the smallest mammal on Earth: it is about the same size as a bumblebee and has an average mass of 2.0 g. Suppose a hog-nosed bat with this mass flies at 2.0 m/s when it detects a bug with a mass of 0.20g flying directly toward it at 8.0 m/s. What fraction of the total kinetic energy
dissipated when it swallows the bug?

Answer 1

• The fraction of the total kinetic energy dissipated is = $0.67$  

Given

$m_1 = 2g\\\\m_2 = 0.2g\\\\v_1 = 2m/s\\\\v_2 = -8m/s$

The expression for a perfectly inelastic collision,

$m_1v_1 + m_2v_2 = (m_1+m_2)v_f$

Therefore,

$v_f = \frac{m_1v_1 + m_2v_2}{(m_1+m_2)} \\\\v_f = \frac{(2*2)+(0.2*-8)}{2+0.2}\\\\v_f = 1.09m/s$

Since the bat and the bug fly towards each other, and after some time the bat swallows the bug, it is said to be a condition of perfectly inelastic collision.

The kinetic energy of the bat before it swallows the bug is,

$KE_b = \frac{1}{2}m_1v_1^2\\\\KE_b = \frac{1}{2}2*2^2\\\\KE_b = 4gm^2/s^2$

The kinetic energy of the bat after it swallows the bug is,

$KE_f = \frac{1}{2}(m_1+m_2)v_f^2\\\\KE_f = \frac{1}{2}(2+0.2)1.09^2\\\\KE_f = 1.3gm^2/s^2$

The total dissipation in the kinetic energy of the bat is calculated as,

$\delta KE = KE_b - KE_f\\\\\delta KE = 4 - 1.3\\\\\delta KE = 2.7 gm^2/s^2$

The fraction of the dissipated kinetic energy of the bat is calculated as,

$d = \frac{\delta KE}{KE_b}\\\\d = \frac{2.7}{4}\\\\d = 0.67$

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