SOMEONE ASKED 👇
asteroid collision?
HERE THE ANSWERS 👇
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you have to use conservation of linear momentum on this problem
momentum (p)=mass x velocity so the original momentum is
p=40m where m is any arbitrary mass
you know that this linear momentum must be conserved after the collision so the combined momentum of the two rocks after should equal this amount. Set up a system of equations using the components of each momentum
x direction
40m=mv(a)cos30+mv(b)cos(-45)
you can cancel the masses in every term and be left with
40=v(a)cos30+v(b)cos(-45)
this is the total momentum in the x-direction.
you need another equation so lets take the y components
it initially has no momentum in the y direction so
0=v(a)sin30+v(b)sin(-45)
so your system of equations looks like this
.5v(a)-.7071v(b)=0
.8660v(a)+.7071v(b)=40
you can add these two equations to solve for v(a) and you get
1.366v(a)=40 so
v(a)=29.3 m/s
plug this in to either equation to get v(b)=
20.7 m/s
for the third part of the question find the kinetic energy of the original asteroid and the kinetic energy of the two final asteroids added together.
so initial kinetic energy is
1/2mv^2=
.5m(40)^2
and final kinetic is
.5m(29.3)^2+.5m(20.7)^2
so you get
800m for initial
and
429.245+214.245=
643.49 for final
you need to find the fraction lost so subtract and get
800-643.49=
156.51
find out what percent of the original this amount is
156.51/800=.196 or
19.6%
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1. There are no “huge” asteroids. The largest asteroid is Ceres, which is 952 km. in diameter. 2. Most asteroids are between the orbts of Mars and Jupiter, but very few get as far out as Jupiter’s orbit itself, so the chances of collision with an asteroid are slim. 3. At least two comets have collided with Jupiter in the past 20 years. They caused temporary scars in Jupiter’s upper atmosphere, but had no lasting effects. 4. The only other large objects n the Solar System are planets, and they are all in stable orbits and never come anywhere close to Jupiter. 5. Jupiter is not large enough to sustain nuclear fusion.
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