Answer to Question #89570 in Astronomy | Astrophysics for Barnabé

Question

A warning system has calculated that two asteroids will collide not far from Earth any time soon.
The smaller asteroid has the mass m and moves with the velocity vm. The bigger asteroid has the mass M = 3m and the velocity of v(M) = 0.5 V(m)
They collide at an angle of α = 60 degree and turn into a
single heavy asteroid (inelastic collision):

(a) Calculate the velocity of the single object after the collision.
(b) Determine the angle Bêta after the collision.

Answer 1

From the conservation of momentum:

$$(m+3m)V\cos{\beta}=3m(0.5v_m)+mv_m\cos{60}$$

$$(m+3m)V\sin{\beta}=mv_m\sin{60}$$

 Thus,

$$(4mV\cos{\beta})^2+(4mV\sin{\beta})^2=(2mv_m)^2+(mv_m\sin{60})^2$$

$$16V^2=v_m^2\left(4+\frac{3}{4}\right)$$

a)

$$V=\frac{\sqrt{19}}{8}v_m$$

b)

$$4m\frac{\sqrt{19}}{8}v_m\sin{\beta}=mv_m\sin{60}$$

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$$\sin{\beta}=\frac{2}{\sqrt{19}}\sin{60}=\sqrt{\frac{3}{19}}$$

$$\beta=\arcsin{\sqrt{\frac{3}{19}}}=23.41\degree$$