Respuesta :
We can use:
2as = v² - u²
to find the deceleration of the meteorite:
a = -u²/2s
a = -(480²)/2 x 0.22
= -5.2 x 10⁵ m/s₂
Force = mass x acceleration
Mass in kg = 27/2.2
Force = 27/2.2 x 5.2 x 10⁵
= 6.4 x 10⁶ Newtons
2as = v² - u²
to find the deceleration of the meteorite:
a = -u²/2s
a = -(480²)/2 x 0.22
= -5.2 x 10⁵ m/s₂
Force = mass x acceleration
Mass in kg = 27/2.2
Force = 27/2.2 x 5.2 x 10⁵
= 6.4 x 10⁶ Newtons
The average force exerted on the meteorite by the car is [tex]\fbox{\begin\\6.4\times{10^6}\,{\text{N}}\end{minispace}}[/tex].
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 [tex]480\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex].
The final speed of meteorite is zero.
The distance covered by meteorite before coming to rest is [tex]22\,{\text{cm}}[/tex].
Mass of the meteorite is [tex]27\,{\text{Pound}}[/tex].
Concept:
Distance covered by meteorite in meter is [tex]0.22\,{\text{m}}[/tex].
Mass of the meteorite in kilograms is [tex]\dfrac{{27}}{{2.2}}\,{\text{kg}}[/tex].
To calculate the acceleration of meteorite when initial speed, final speed, and distance covered given, the equation of motion used is:
[tex]\fbox{\begin\\{v^2} = {u^2} + 2as\end{minispace}}[/tex]
Rearrange the above equation for acceleration:
[tex]a = \dfrac{{{v^2} - {u^2}}}{{2s}}[/tex]
Here, [tex]a[/tex] is the acceleration of meteorite, [tex]u[/tex] is the initial speed, [tex]v[/tex] is the final speed and [tex]s[/tex] is the distance covered by meteorite before coming to rest.
Substitute [tex]0\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]v[/tex] , [tex]480\text{ m}/\text{s}[/tex] for [tex]u[/tex] and [tex]0.22\text{ m}[/tex] for [tex]s[/tex] in above equation.
[tex]\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-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\hfill\\\end{aligned}[/tex]
Equation for the average force exerted by car on the meteorite:
[tex]\fbox{F = ma}[/tex]
Here, [tex]F[/tex] is the force, [tex]m[/tex] is the mass and [tex]a[/tex] is the acceleration of meteorite.
Substitute [tex]\dfrac{{27}}{{2.2}}\,{\text{kg}}[/tex] for [tex]m[/tex] and [tex]-5.2\times{10^5}\text{ m}/\text{s}^2[/tex] for [tex]a[/tex] in above equation.
[tex]\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-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}\right)\hfill\\&=-6.4\times{10^6}\,{\text{N}}\hfill\\\end{aligned}[/tex]
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 [tex]\fbox{\begin\\6.4\times{10^6}\,{\text{N}}\end{minispace}}[/tex].
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.
