Question

A $300\mathrm{g}$bird flying along at $6.0\mathrm{m}/\mathrm{s}$sees a $10\mathrm{g}$ insect heading straight toward it at a speed of $30\mathrm{m}/\mathrm{s}$. The bird opens its mouth wide and enjoys a nice lunch. What is the bird’s speed immediately after swallowing?

Short answer

The final speed of the bird is$4.8\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Given information

We have given,

Speed of bird =$6\mathrm{m}/\mathrm{s}$

Speed of insect =$30\mathrm{m}/\mathrm{s}$

Mass of bird =$300\mathrm{g}$

Mass of insect =$10\mathrm{g}$

We have to find the bird's final speed after he eat the insect.

Simplify

We know from the conservation of momentum that in the colleen the momentum before the Collison of the system will be same after the Collison.

momentum before Collison = Momentum after Collison

${\mathrm{P}}_{\mathrm{I}}={\mathrm{P}}_{\mathrm{F}}$

since initial momentum will be the sum of momentums of the bird and insect individuals.

$$\begin{gathered} {\mathrm{P}}_{\mathrm{I}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{V}}_{\mathrm{B}}+{\mathrm{m}}_{\mathrm{I}}+{\mathrm{V}}_{\mathrm{I}} \\ {\mathrm{P}}_{\mathrm{I}}=0.3\mathrm{kg}\times 6\mathrm{m}/\mathrm{s}+0.01\mathrm{kg}\times -30\mathrm{m}/\mathrm{s} \\ {\mathrm{P}}_{\mathrm{F}}=(0.3+0.01)\mathrm{kg}\times \mathrm{V} \end{gathered}$$

THEN,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{F}}={\mathrm{P}}_{\mathrm{I}} \\ 0.31\mathrm{kg}\times \mathrm{V}=(0.18-0.3) \\ \mathrm{V}=4.8\mathrm{m}/\mathrm{s} \end{gathered}$$