Question

Locusts can jump to heights of 0.870 m.

The locust actually jumps at an angle of about 55.0° to the horizontal, and air resistance is not negligible. The result is that the takeoff speed is about 40.0% higher than the takeoff speed when the locust jumps straight up, and ignoring air resistance. If the mass of the locust is 2.00 g and its body moves 4.00 cm in a straight line while accelerating from rest to the takeoff speed, calculate the acceleration of the locust (assumed constant).

__________ m/s²

The locust actually jumps at an angle of about 55.0° to the horizontal, and air resistance is not negligible. The result is that the takeoff speed is about 40.0% higher than the takeoff speed when the locust jumps straight up, and ignoring air resistance. If the mass of the locust is 2.00 g and its body moves 4.00 cm in a straight line while accelerating from rest to the takeoff speed, estimate the force exerted on the hind legs by the ground (Ignore the locust’s weight).

_____N

The locust actually jumps at an angle of about 55.0° to the horizontal, and air resistance is not negligible. The result is that the takeoff speed is about 40.0% higher than the takeoff speed when the locust jumps straight up, and ignoring air resistance. If the mass of the locust is 2.00 g and its body moves 4.00 cm in a straight line while accelerating from rest to the takeoff speed, compare the force exerted on the hind legs by the ground with the locust’s weight.

The force exerted on the hind legs by the ground is ________ times that of the locust’s weight.

Answer 1

Here we have given that the maximum height to which a locust can jump is 0.870 m

Lets first consider the locust's motion if it jumps straight up, ignoring air resistance.

The velocity-displacement relation is given as,

v^2=u^2+2as....(1)

If the locust leaps straight up with an initial velocity u, then on reaching a vertical displacement of

s=0.870m so its velocity becomes zero

(g=9.81 m/s2 )

So the velocity-displacement relation becomes:

0=u^2−2×9.81×0.87

or u=√1.74×9.8= 4.1315 m/s

Also we have given that the locust actually makes an angle with the horizontal when it lifts off and resistance influences the motion.

Also given that the take-off speed when we take these real-life situations into consideration is 40% higher than the take-off speed if the locust jumps straight up and if we ignore air resistance.

Thus in real life, the take off speed of the locust is

u' = 1.4×u =1.4×4.1315=5.78411m/s

The mass of the locust is

m=2.00 g=0.002 kg

Given that in real-life the body of the locust moves in a straight line through a distance

s=4.00 cm =0.04 m

accelerating from rest to the take off speed.

Then using equation (1) we have the condition,

5.78411^2=0 + 2×a×0.04

A.

The acceleration of the locust during lift off is therefore a=5.78411^2/(2×0.04)=418.200 m/s2

B.

The force exerted by the ground on the hindlegs of the locust during take off is then :

F=ma=0.002×418.2003=0.8364006 N

C.

The locust's weight

mg=0.002×9.81=0.01962 N

Thus the force exerted on the hind legs by the ground is 42.63 times that of the locust’s weight.