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Sergeeva-Olga [200]
4 years ago
5

Two asteroids collide and stick together. The first asteroid has mass of 1.50 × 104 kg andis initially moving at 0.77 × 103 m/s.

The second asteroid has mass of 2.00 × 104 kg and ismoving at 1.02× 103 m/s. Their initial velocities made an angle of 20° with respect to eachother. What is the final speed and direction with respect to the velocity of the first asteroid?
Physics
1 answer:
KengaRu [80]4 years ago
7 0

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:

p_{1ox}+p_{2ox}=p_{fx}\\\\p_{2oy}=p_{fy}

And, since p=mv, we get:

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}

Solving for v_fx and v_fy, and calculating their values, we get:

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

Now, the final speed can be calculated using the Pythagorean Theorem:

v_f=\sqrt{v_{fx}^{2}+v_{fy}^{2}} \\\\\implies v_f=\sqrt{(878m/s)^{2}+(199m/s)^{2}}=900m/s

And the direction \beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°can be obtained using trigonometry:

\beta=\arctan \frac{v_{fy}}{v_{fx}}\\ \\\implies \beta=\arctan\frac{199m/s}{878m/s}=12.8\°

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.

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Explanation:

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\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}.

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\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}.

Calculate v(\text{earth, after}) using m(\text{earth}) and v(\text{athlete, after}) values from the question:

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The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately 2.0 \times 10^{-23}\; \rm m \cdot s^{-1}.

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