Answer:
Final speed is 900.06 m/s at
Solution:
As per the question:
Mass of the first asteroid, m = ![15\times 10^{3}\kg](https://tex.z-dn.net/?f=15%5Ctimes%2010%5E%7B3%7D%5Ckg)
Mass of the second asteroid, m' = ![20\times 10^{3}\kg](https://tex.z-dn.net/?f=20%5Ctimes%2010%5E%7B3%7D%5Ckg)
Initial velocity of the first asteroid, v = 770 m/s
Initial velocity of the second asteroid, v' = 1020 m/s
Angle between the two initial velocities, ![\theta = 20^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2020%5E%7B%5Ccirc%7D)
Now,
Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:
![\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}](https://tex.z-dn.net/?f=%5Cvec%7BR%7D%20%3D%20%5Csqrt%7BA%5E%7B2%7D%20%2B%202ABcos%5Ctheta%20%2B%20B%5E%7B2%7D%7D)
Thus applying vector addition and momentum conservation, the final velocity is given by:
(1)
Now,
![(m +m')v_{final} = (35\times 10^{3})v_{final}](https://tex.z-dn.net/?f=%28m%20%2Bm%27%29v_%7Bfinal%7D%20%3D%20%2835%5Ctimes%2010%5E%7B3%7D%29v_%7Bfinal%7D)
![(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}](https://tex.z-dn.net/?f=%28mv%29%5E%7B2%7D%20%3D%20%2815%5Ctimes%2010%5E%7B3%7D%5Ctimes%20770%29%5E%7B2%7D%20%3D%201.334%5Ctimes%2010%5E%7B14%7D)
![(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}](https://tex.z-dn.net/?f=%28m%27v%27%29%5E%7B2%7D%20%3D%20%2820%5Ctimes%2010%5E%7B3%7D%5Ctimes%201020%29%5E%7B2%7D%20%3D%204.16%5Ctimes%2010%5E%7B14%7D)
![2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}](https://tex.z-dn.net/?f=2%28mv%29%28m%27v%27%29cos20%5E%7B%5Ccirc%7D%20%3D%202%2815%5Ctimes%2010%5E%7B3%7D%5Ctimes%20770%29%2820%5Ctimes%2010%5E%7B3%7D%5Ctimes%201020%29cos20%5E%7B%5Ccirc%7D%20%3D%204.43%5Ctimes%2010%5E%7B14%7D)
Now, substituting the suitable values in eqn (1), we get:
![v_{final} = 900.06\ m/s](https://tex.z-dn.net/?f=v_%7Bfinal%7D%20%3D%20900.06%5C%20m%2Fs)
Now, the direction for the two vectors is given by:
![\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20sin%5E%7B-%201%7D%20%5Cfrac%7Bm%27v%27sin20%5E%7B%5Ccirc%7D%7D%7B%28m%20%2B%20m%27%29v_%7Bfinal%7D%7D)
![\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20sin%5E%7B-%201%7D%20%5Cfrac%7B20%5Ctimes%2010%5E%7B3%7D%5Ctimes%201020sin20%5E%7B%5Ccirc%7D%7D%7B%2835%5Ctimes%2010%5E%7B3%7D%29%5Ctimes%20900.06%7D%20%3D%200.2215%5E%7B%5Ccirc%7D)