Respuesta :
Answer:
Magnitude 900m/s, direction 12.8° respect to the velocity of the first asteroid.
Explanation:
This is a perfectly inelastic collision, because the two asteroids stick together at the end. That means that the kinetic energy doesn't conserves, but the linear momentum does. But, since the velocities of the asteroids have different directions, we have to break down them in components. For convenience, we will take the direction of the first asteroid as x-axis, and its perpendicular direction (in the plane of the two velocity vectors) as y-axis. So, we have that:
[tex]p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}[/tex]
And, since [tex]p=mv[/tex], we get:
[tex]m_1v_{1o}+m_2v_{2o}\cos\theta=(m_1+m_2)v_{fx}\\\\m_2v_{2o}\sin\theta=(m_1+m_2)v_{fy}[/tex]
Solving for v_fx and v_fy, and calculating their values, we get:
[tex]v_{fx}=\frac{m_1v_{1o}+m_2v_{2o}\cos\theta}{m_1+m_2}\\\\\implies v_{fx}=\frac{(1.50*10^{4}kg)(0.77*10^{3}m/s)+(2.00*10^{4}kg)(1.02*10^{3}m/s)\cos20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=878m/s\\\\\\v_{fy}=\frac{m_2v_{2o}\sin\theta}{m_1+m_2}\\\\\implies v_{fy}=\frac{(2.00*10^{4}kg)(1.02*10^{3}m/s)\sin20\°}{1.50*10^{4}kg+2.00*10^{4}kg}=199m/s[/tex]
Now, the final speed can be calculated using the Pythagorean Theorem:
[tex]v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s[/tex]
And the direction [tex]\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°[/tex]can be obtained using trigonometry:
[tex]\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°[/tex]
That means that the final velocity of the two asteroids has a magnitude of 900m/s and a direction of 12.8° with respect to the velocity of the first asteroid.
The final speed is 900 m/s and the angle with the velocity of the first asteroid is 7.2°.
Conservation of momentum:
Given information:
Masses of asteroids,
m = 1.5 × 10⁴ kg
M = 2 × 10⁴ kg
and their initial velocities are:
u = 0.77 × 10³ m/s
U = 1.02 × 10³ m/s
Let u makes an angle of θ = 20° with respect to U.
Let v be the final speed of the asteroids after they collide and move together, which is an example of inelastic collision.
In an inelastic collision, the momentum remains conserved:
momentum in the x-direction or horizontal direction:
mucosθ + MU = (m + M)vₓ
(1.5 × 10⁴)×( 0.77 × 10³) cos20 + (2 × 10⁴)( 1.02 × 10³ ) = (1.5 × 10⁴ + 2 × 10⁴)vₓ
vₓ = 893 m/s
momentum in y- direction:
musinθ = (m + M)v[tex]__y[/tex]
(1.5 × 10⁴)×( 0.77 × 10³) sin20 = (1.5 × 10⁴ + 2 × 10⁴)v[tex]_y[/tex]
v[tex]_y[/tex] = 113 m/s
So the resultant speed:
[tex]v = \sqrt{v_x^2+v_y^2}\\\\v=\sqrt{893^2+113^2}[/tex]
v = 900 m/s
The angle with respect to velocity of first asteroid is:
Ф = tan⁻¹(113/893
Ф = 7.2°
Learn more about conservation of momentum:
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