Asteroid of mass 10^16 kg is travelling in the equatorial plane of Earth. It hits the surface at 45°. After the impact the day shortens by 1% (15 mins). How fast was the asteroid - comment? Neglect effects of atmosphere. Consider only inelastic collision.

Key idea - total angular momentum of the system is conserved. Before the collision: angular momentum of earth is given by L_E=Iw_E, here w_E is the angular velocity of Earth and I is its moment of innertia which is given by I=2/5M_ER². R is the radius of Earth and M_E is its mass. Angular momentum of the asteroid is L_A=mvd. Here m is asteroid's mass, v its velocity and d the perpendicular distance of its trajectory from the centre of the Earth given by d=sin(45°)R (see the whiteboard).
After the collision (inelastic) the asteroid is stuck to the surface and the whole system rotates at different angular velocity w_F. Total angular momentum is therefore given by L_F=Iw_F + mR²w_F.
By equating the two angular momenta we obtain an equation from which we can express v - the asteroid's velocity. By neglecting the asteroids mass compared to the mass of the Earth (factor of 10⁸) we obtain an approximate expression v=(I(w_F-w_E))/(m*sin(45°)R) = 15.910^8 i.e. faster than the speed of light!! This means that our classical approximations break down and we need to treat the whole problem relativistically which is well beyond the scope of A level Physics.