The average force exerted on the meteorite by the car is \(\fbox{\begin\\6.4\times{10^6}\,{\text{N}}\end{minispace}}\).
Further Explanation:
The meteorite hitting the car followed the law of motion. Newton’s equation of motion applied to define the average force on meteorite.
Given:
The initial speed of meteorite is \(480\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-
ulldelimiterspace} {\text{s}}}\).
The final speed of meteorite is zero.
The distance covered by meteorite before coming to rest is \(22\,{\text{cm}}\).
Mass of the meteorite is \(27\,{\text{Pound}}\).
Concept:
Distance covered by meteorite in meter is \(0.22\,{\text{m}}\).
Mass of the meteorite in kilograms is \(\dfrac{{27}}{{2.2}}\,{\text{kg}}\).
To calculate the acceleration of meteorite when initial speed, final speed, and distance covered given, the equation of motion used is:
\(\fbox{\begin\\{v^2} = {u^2} + 2as\end{minispace}}\)
Rearrange the above equation for acceleration:
\(a = \dfrac{{{v^2} - {u^2}}}{{2s}}\)
Here, \(a\) is the acceleration of meteorite, \(u\) is the initial speed, \(v\) is the final speed and \(s\) is the distance covered by meteorite before coming to rest.
Substitute \(0\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-
ulldelimiterspace} {\text{s}}}\) for \(v\) , \(480\text{ m}/\text{s}\) for \(u\) and \(0.22\text{ m}\) for \(s\) in above equation.
\(\begin{aligned}a&=\dfrac{(0\text{ m}/s)^2-(480\text{ m}/{s})^2}{2\times0.22\text{ m}}\\&=-5.2\times{10^5}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-
ulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\hfill\\\end{aligned}\)
Equation for the average force exerted by car on the meteorite:
\(\fbox{F = ma}\)
Here, \(F\) is the force, \(m\) is the mass and \(a\) is the acceleration of meteorite.
Substitute \(\dfrac{{27}}{{2.2}}\,{\text{kg}}\) for \(m\) and \(-5.2\times{10^5}\text{ m}/\text{s}^2\) for \(a\) in above equation.
\(\begin{aligned}F&=\left({\dfrac{{27}}{{22}}\,kg} \right)\times\left({-5.2\times{{10}^5}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-
ulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}\right)\hfill\\&=-6.4\times{10^6}\,{\text{N}}\hfill\\\end{aligned}\)
Here, the negative sign indicates that the force is opposite to the direction of motion of meteorite.
Thus, the average force exerted on the meteorite by the car is \(\fbox{\begin\\6.4\times{10^6}\,{\text{N}}\end{minispace}}\).
Learn more:
1. Conservation of momentum during collision https://brainly.com/question/9484203
2. Two dimensional motion of a ball https://brainly.com/question/11023695
3. A 30 kg box pulled across a carpeted floor https://brainly.com/question/7031524
Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Meteorite, October 9, 1992, pound, initial speed, dent, 22cm deep, average force, and two significant figures, force, acceleration, car.
