Question

A small $10^{9}kg$ asteroid, moving in the positive x direction at 10,000 m/s, collides with a $10^{8}kg$ asteroid moving in the positive y direction at 20,000 m/s. The two stick together and form a single object.

What is the velocity, and angle of the pair after the collision?

Answer 1

using momentum conservation:

In the x axis:

(I)

using momentum conservation: In the x axis: m_ 1 v_ 2 = (m_ 1 + m_ 2) v cos theta (I) In the y axis: m_ 2 v_ 2 = (m_ 1 + m_ 2) v sin theta (II) II/I tan theta = m_ 2 V_ 2/m_ 1 v_ 1 = 10^8 middot 20000/10^9 middot 10000 theta = 11.30 degree plugging the angle in any of the equations: m_ 1 v_ 1 = (m_ 1 + m_ 2) v cos theta 10^9 middot 10000 = (10^9 + 10^8) v cos 11.3 solving for v v = 9270.62 m/s

In the y axis:

$m_2v_2=(m_1+m_2)vsin\theta$   (II)

(II)/(I)

$tan\theta=\frac{m_2v_2}{m_1v_1}=\frac{10^{8}\cdot 20000}{10^{9}\cdot 10000}$

$\theta=11.30^{\circ}$

Plugging the angle in any of the equations:

using momentum conservation: In the x axis: m_ 1 v_ 2 = (m_ 1 + m_ 2) v cos theta (I) In the y axis: m_ 2 v_ 2 = (m_ 1 + m_ 2) v sin theta (II) II/I tan theta = m_ 2 V_ 2/m_ 1 v_ 1 = 10^8 middot 20000/10^9 middot 10000 theta = 11.30 degree plugging the angle in any of the equations: m_ 1 v_ 1 = (m_ 1 + m_ 2) v cos theta 10^9 middot 10000 = (10^9 + 10^8) v cos 11.3 solving for v v = 9270.62 m/s

$10^{9}\cdot 10000=(10^{9}+10^{8})vcos11.3$

solving for v