Question

A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms; these are all typical values. at what speed will it leave the ground?

Answer 1

Answer:

v = 0.509 m/s

Explanation:

Given that,

Mass of the ant, $m = 12\ mg = 1.2\times 10^{-5}\ kg$

Force exerted, $F = 47\ mN = 0.047\ N$

Time, $t=0.13\ ms=0.00013\ s$

To find,

Speed with which it leave the ground.

Solution,

The change in momentum is equal to the product of force and time as :

$F\times t=m(v-u)$

Since, u = 0

$F\times t=mv$

$v=\dfrac{Ft}{m}$

$v=\dfrac{0.047\ N\times 0.00013\ s}{1.2\times 10^{-5}\ kg}$

v = 0.509 m/s

So, the speed with which it will leave the ground is 0.509 m/s.