Problem help...?

Question

During an autumn storm, a 0.012 kg hail stone traveling at 20 m/s made a 0.2 cm deep dent in the hood fo a car. What average force did the car exert to stop the hail stone?

Please show how you did it, equations, etc.

Answer 1

Apply the law of conservation of energy, i.e.,

Kinetic energy of hail stone = energy absorbed by car hood

(1/2)mv^2 = Fd

were

m = mass of the hail stone = 0.012 kg (given)

v = hail stone velocity = 20m/sec (given)

F = force exerted by car hood on the hail stone

d = car hood dent caused by the falling stone = 0.2 cm = 0.002 m (given)

Substituting appropriate values,

(1/2)(0.012)(20)^2 = F(0.002)

Solving for "F",

F = (1/2)(0.012)(400)/0.002

F = 1200 Newtons

Answer 2

Hail stone travelling at 20 m/s stopped after a distance of 0.2 cm (=0.002 m). If its deceleration is a,

(20)^2 = 2a*(0.002)

=> a = 10^5 m/s^2

=> Force = mass x deceleration

= 0.012 x 10^5 N

= 1200 N.