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IRINA_888 [86]
3 years ago
9

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

ing or eating–joyfully scampering about on the cage\'s floor. Bryce tracks his mice\'s health diligently and just now recorded their masses as 0.0225 kg, 0.0223 kg, 0.0197 kg, and 0.0127 kg. At this very instant the x and y components of the mice\'s velocities are, respectively, (0.869 m/s, -0.283 m/s), (-0.883 m/s, -0.253 m/s), (0.345 m/s, 0.803 m/s), and (-0.555 m/s, 0.205 m/s). Calculate the x and y components of Bryce\'s mice\'s total momentum, px and py.
Physics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

  m1 = 0.0225 Kg

  V1 = (0.869, -0.283) m / s

 

  Px = m Vx

  Px1 = 0.0225 0.869

  Px1 = 0.01955 Kg m

  Py = m Vy

  Py1 = 0.0225 (-0.283)

  Py1 = -0.006368 Kg m

  P1 = (0.0196, -0.00637) Kg m

Mouse 2

 m2 = 0.0223 Kg

 Px2 = 0.0223 (-0.883) = -0.0196 Kg m

 Py2 = 0.0223 (-0.253) = -0.00564 Kg m

 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

 m3 = 0.0197

 Px3 = 0.0197 0.345 = 0.00680 Kg m

 Py3 = 0.0197 0.803 = 0.0158 Kg m

 P3 = (0.00680, 0.0158) Kg m

Mouse4

  m4 = 0.0127 Kg

  Px4 = 0.0127 (-0.555) = -0.00705 Kg m

  Py4 = 0.0127 0.205 = 0.00260 Kg m

  P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

   Px = Px1 + Px2 + Px3 + Px4  

   Py = py1 + Py2 + Py3 + Py4

   Px = 0.0196 -0.0196 +0.00680 -0.00705

   Px = -0,000250 Kg m

   Py = -0.00637 -0.00564 +0.0158 +0.00260

   Py = 0.00639 Kg m

   P = (-0.000250, 0.00639) Kg m

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3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

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Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

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c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

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