Question

Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight down at a mass rate of m/?t = 0.030 kg/s and an initial velocity of v0 = -15 m/s and strikes the roof perpendicularly. Suppose that the hail bounces off the roof of the car with a velocity of +15 m/s. Ignoring the weight of the hailstones, calculated the force exerted by the hail on the roof. __________N

Answer 1

0.9 N

Step-by-step explanation:

The force exerted on an object is related to its change in momentum by:

`F=(\Delta p)/(\Delta t)`

where

F is the force exerted

`\Delta p` is the change in momentum

`\Delta t` is the time interval

The change in momentum can be rewritten as

`\Delta p = m(v-u)`

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

`F=(m(v-u))/(\Delta t)`

In this problem we have:

`(m)/(\Delta t)=0.030 kg/s` is the mass rate

`u=-15 m/s` is the initial velocity

`v=+15 m/s` is the final velocity

Therefore, the force exerted by the hail on the roof is:

`F=(0.030)(+15-(-15))=0.9 N`