Question

Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially moving at 770 \, \mathrm{m/s}770m/s. The second asteroid has a mass of 20 \times 10^3\,\mathrm{kg}20×10 3 kg and is moving at 1020\,\mathrm{m/s}1020m/as. Their initial velocities made an angle of 20^\circ20 ∘ with respect to each other. What is the final speed and direction with respect to the velocity of the first asteroid?

Answer 1

Final speed is 900.06 m/s at
`0.2215^(\circ)`

Solution:

As per the question:

Mass of the first asteroid, m =
`15* 10^(3)\kg`

Mass of the second asteroid, m' =
`20* 10^(3)\kg`

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)`

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)}`

Thus applying vector addition and momentum conservation, the final velocity is given by:

`(m + m')v_(final) = \sqrt{(mv)^(2) + 2(mv)(m'v')cos20^(\circ) + (m'v')^(2)}` (1)

Now,

`(m +m')v_(final) = (35* 10^(3))v_(final)`

`(mv)^(2) = (15* 10^(3)* 770)^(2) = 1.334* 10^(14)`

`(m'v')^(2) = (20* 10^(3)* 1020)^(2) = 4.16* 10^(14)`

`2(mv)(m'v')cos20^(\circ) = 2(15* 10^(3)* 770)(20* 10^(3)* 1020)cos20^(\circ) = 4.43* 10^(14)`

Now, substituting the suitable values in eqn (1), we get:

`v_(final) = 900.06\ m/s`

Now, the direction for the two vectors is given by:

`\theta = sin^(- 1) (m'v'sin20^(\circ))/((m + m')v_(final))`

`\theta = sin^(- 1) (20* 10^(3)* 1020sin20^(\circ))/((35* 10^(3))* 900.06) = 0.2215^(\circ)`