Question

Meteorite On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, creating a dent about 22 cm deep. If the initial speed of the meteorite was 550 m/s, what was the average force exerted on the meteorite by the car?

Answer 1

The average force exerted on the meteorite by the car is 8.435x10⁶N

Step-by-step explanation:

Given:

m = 27 lb = 12.27 kg

D = 22 cm = 0.22 m

v1 = 550 m/s

The acceleration of the meteorite is:

`v_(2) ^(2) =v_(1) ^(2)+2aD\\0=550^(2) +2a(0.22)\\a=-(550^(2) )/(2*0.22) =-687500m/s^(2)`

Negative sign indicates opposite to the initial velocity. The average force is equal to:

Magnitude of a = 687500 m/s²

F = m*a = 12.27 * 687500 = 8.435x10⁶N

Answer 2

8427375 N

Step-by-step explanation:

From newton's equation of motion,

F = ma................. Equation 1

Where F = average force exerted on the meteorite by the car, m = mass of the meteorite, a = acceleration.

We can find the value of acceleration from newton's equation of motion

using,

v² = u²+2as.................. Equation 2

Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

Make a the subject of the equation

a = (v²-u²)/2s............... Equation 3

Note: When the meteorite struck the car, it was brought to rest

Given: u = 550 m/s, v = 0 m/s ( brought to rest), s = 22 cm = 0.22 m

Substitute into equation 3

a = (0²-550²)/(2×0.22)

a = -302500/0.44

a = -687500 m/s²

Also given: m = 27 pound = (27×0.454) = 12.258 kg

Substitute into equation 1

F = 12.258(-687500)

F = -8427375 N

Note: The negative sign tells that the force act in opposite direction to the motion of the meteorite