Question #89570
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.
1
Expert's answer
2019-05-15T10:11:56-0400

From the conservation of momentum:


(m+3m)Vcosβ=3m(0.5vm)+mvmcos60(m+3m)V\cos{\beta}=3m(0.5v_m)+mv_m\cos{60}

(m+3m)Vsinβ=mvmsin60(m+3m)V\sin{\beta}=mv_m\sin{60}

 Thus,


(4mVcosβ)2+(4mVsinβ)2=(2mvm)2+(mvmsin60)2(4mV\cos{\beta})^2+(4mV\sin{\beta})^2=(2mv_m)^2+(mv_m\sin{60})^2


16V2=vm2(4+34)16V^2=v_m^2\left(4+\frac{3}{4}\right)

a)


V=198vmV=\frac{\sqrt{19}}{8}v_m

b)


4m198vmsinβ=mvmsin604m\frac{\sqrt{19}}{8}v_m\sin{\beta}=mv_m\sin{60}

​​

sinβ=219sin60=319\sin{\beta}=\frac{2}{\sqrt{19}}\sin{60}=\sqrt{\frac{3}{19}}

β=arcsin319=23.41°\beta=\arcsin{\sqrt{\frac{3}{19}}}=23.41\degree



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