Asteroid colliding with a planet

[henry3369]

Homework Statement

Suppose that an asteroid traveling straight toward the center of the Earth were to collide with our planet at the equator and bury itself just below the surface.

What would have to be the mass of this asteroid, in terms of the Earth's mass M , for the day to become 26.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the Earth and that the Earth is uniform throughout.

Homework Equations

Conservation of angular momentum: L_initial = L_final

The Attempt at a Solution

ω_final = (0.74)ω_initial (This makes the final speed 26% slower than the original)L_initial = L_final
I_initialω_initial = I_finalω_final
I_initialω_initial = I_final(0.74)ω_initial (ω_initial cancel out)
(2/5)MR² = ((2/5)MR² + mR²)(0.74) (R²'s cancel out)
(2/5)M = ((2/5)M + m) (0.74) (Distribute the 0.74)
(2/5)M = ((1.48/5)M + 0.74m)
0.74 m = (2/5)M - (1.48/5)M
0.74m = M((2/5) - (1.48/5))
0.74m = 0.104M
m = 0.141M

This answer is m = 0.104M which is on the right side of the second to last line (0.74m = 0.104M), but I still have the 0.74 on the left side. Where is my mistake?

 

[gneill]

The rotation period is given by $\frac{2 \pi}{ω}$ for whatever the current value of ω is. For the new period to be 26% larger, then
$$\frac{2 \pi}{ω_f} = 1.26 \frac{2 \pi}{ω_o} $$