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 vM =
1
2
vm. They collide at an angle of α = 60◦ and turn into a
single heavy asteroid (inelastic collision):
M
m
α
β
(a) Calculate the velocity of the single object aer the collision.
(b) Determine the angle β aer 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,

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

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

a)

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

b)

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

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

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