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sladkih [1.3K]
2 years ago
15

A gadget of mass 21.85 kg floats in space without motion. Because of some internal malfunction, the gadget violently breaks up i

nto 3 fragments flying away from each other. The first fragment has mass m1 = 6.42 kg and speed v1 = 6.8 m/s while the second fragment has mass m2 = 8.26 kg and speed v2 = 3.54 m/s. The angle between the velocity vectors ~v1 and ~v2 is θ12 = 64 ◦ . What is the speed v3 of the third fragment? Answer in units of m/s.

Physics
1 answer:
BaLLatris [955]2 years ago
6 0

Answer:

v_3=8.622\ m.s^{-1}

Explanation:

Given:

  • mass of the gadget, m=21.85\ kg
  • mass of fragment 1, m_1=6.42\ kg
  • mass of fragment 2, m_2=8.26\ kg
  • velocity of fragment 1, v_1=6.8\ m.s^{-1}
  • velocity of fragment 2, v_2=3.54\ m.s^{-1}
  • angle between the velocity 1 & 2, \theta=64\ ^{\circ}

<u>Now the mass of the third part:</u>

m_3=m-(m_1+m_2)

m_3=21.85-(6.42+8.26)

m_3=7.17\ kg

<u>The momentum of the gadget before and after collision is equal:</u>

m_3.v_3=m_1.v_1\cos\ \frac{\theta}{2} +m_2.v_2\cos \frac{\theta}{2}

7.17\times v_3=(6.42\times6.8+8.26\times3.54)\times \cos \frac{64}{2}

v_3=8.622\ m.s^{-1}

Refer schematic for more clarity.

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Answer:

(a)

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\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

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1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

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h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

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To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

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