Question

A 0.3 g mosquito is flying toward a girl with a speed of 4.5 mph. Just before landing on the girl, the fly is swatted straight back at a speed of 12 mph. If the fly swatter and the fly were in contact for 0.2 s, what is the force that was exerted on the fly

Answer 1

Answer:

1.1x10^-2N

Explanation:

We have the change in momentum as

P = 0.3(4.5+12)g.mph

= 0.3x0.447x(4.5+12)x10^-3

Then the force that is exerted will be

F = p/∆t

∆t = 0.2

= 0.3x0.447x(4.5+12)x10^-3/0.2

= 0.1341x16.5x10^-3/0.2

= 1.1x10^-2

Therefore the force that was exerted is equal to 1.1x10^-2