Question

A warning system has calculated that two asteroids will collide not far from Earth any time soon.
The smaller asteroid has the massmandmoves with the velocity vm. The bigger asteroid has the
massM = 3mand the velocity of vM = 1
2vm. They collide at an angle of  = 60 and turn into a
single heavy asteroid (inelastic collision):
M
m

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

Answer 1

$$M=3m, v_M={1 \over 2}v_m, \alpha=60^\circ$$

In inelastic collisions the kinetic energy of the system of objects is not conserved after the collision.

According the conservation of momentum: 

$$m\overrightarrow{v_m}+M\overrightarrow{v_M}=(m+M)\overrightarrow{u}$$

In projections on the coordinate axes

$$mv_m+Mv_M\cos\alpha=(m+M)u\cos \theta$$

$$0+Mv_M\sin \alpha=(m+M)u\sin \theta$$

$$\tan \theta={Mv_M\sin \alpha \over mv_m+Mv_M\cos\alpha}$$

Substitute

$$\tan \theta={3m({1 \over 2}v_m)\sin 60^\circ \over mv_m+3m({1 \over 2}v_m)\cos60^\circ}={3\sqrt{3} \over 7}$$

$$\theta=\arctan({3\sqrt{3} \over 7})\approx36.6^\circ$$

$$mv_m+3m({1 \over 2}v_m)\cos60^\circ=(m+3m)u\sin (\arctan({3\sqrt{3} \over 7}))$$

$$u={7 \over 16\sin (\arctan({3\sqrt{3} \over 7}))}v_m\approx 0.734v_m$$

(a)

$$u={7 \over 16\sin (\arctan({3\sqrt{3} \over 7}))}v_m\approx 0.734v_m$$

(b)

The angle is

$$\theta=\arctan({3\sqrt{3} \over 7})\approx36.6^\circ$$

with respect to initial direction of the velocity of asteroid m.