Two asteroids collide in space. The collision is not completely inelastic--the a

Question

Two asteroids collide in space. The collision is not completely inelastic--the asteroids pull apart and continue on their way. Their masses are 8.7 X 108 kg for Asteroid A and 3.4 X 108 kg for Asteroid B. Treat Asteroid A as being initially at rest. Asteroid B, moving at high speed, strikes Asteroid A and moves from the collision at a speed of 7300 m/s in a direction 48 degrees from its original path. Meanwhile, Asteroid A moves away from the collision at a spee of 4400 m/s at an angle of 27 degrees from the original path of Asteroid B, but in the opposite direction, of course. Assume no significant amount of mass is lost in the collision (a definite oversimplification).

a) Find the initial momentum, and from it the original speed, of Asteroid B before the collision.

b) How much kinetic energy was lost in the collision?

Explanation / Answer

Per conservation of momentum, mv = mv_a + mv_b, where m is the equal mass, v the initial velocity of A, and v_a and v_b are the final velocities of A and B. We may choose a frame of reference so that the initial velocity of A is along the x-axis.

Since all the masses are the same, we can divide by m and we get that the vector sum of the final velocities must be equal to the initial velocity -- since we have chosen a reference frame where the initial velocity has only an x-component, it must be that the y components of the velocities cancel:

v_a sin(30) = v_b sin(45)
v_a(1/2) = v_b [sqrt(2)/2]
v_a = v_b sqrt(2)

Also, the x-components must add to the initial velocity v:

v_a cos(30) + v_b cos(45) = v
v_a [sqrt(3)/2] + v_b [sqrt(2)/2] = v

From the above:
[v_b sqrt(2)][sqrt(3)/2] + v_b [sqrt(2)/2] = v
v_b[(sqrt(6)+sqrt(2))/2] = v --> v _b = 2v/[sqrt(6) + sqrt(2)]

Then v_a = v_b sqrt(2) = 2v sqrt(2)/[sqrt(6) + sqrt(2)]

I am writing this out at work without the benefit of a calculator, so I don't have the luxury of computing the response to C, but you need only find the initial KE_i of 1/2*m*v^2, and the final KE_f of 1/2*m*v_a^2 + 1/2*m*v_b^2 and compute 1 - KE_f/KE_i.

Edit: Pulled up the calculator in Windows, and my velocities agree with you. Since there are 1/2m's present in every term, you can cancel them out again and you have the ratio of final to initial KE as: [(29.3)^2+(20.7)^2]/40^2 = 80.44%, so there is 19.56% loss in the collision.

ans 2

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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