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alexira [117]
3 years ago
11

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

/s.After the collision, ~vA = (−5.0ˆi + 20ˆj) m/s. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
Physics
1 answer:
AleksandrR [38]3 years ago
6 0

Answer:

Part a)

10\hat i + 15\hat j = \vec v

Part b)

\Delta K = 500 J

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i  + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v

5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v

10\hat i + 15\hat j = \vec v

Part b)

magnitude of the initial speed of A = \sqrt{15^2 + 30^2} = 33.54 m/s

magnitude of the initial speed of B = \sqrt{10^2 + 5^2} = 11.18 m/s

magnitude of final speed of A = \sqrt{5^2 + 20^2} = 20.61 m/s

magnitude of final speed of B = \sqrt{10^2 + 15^2} = 18.03 m/s

Now change in total kinetic energy is given as

\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)

\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)

\Delta K = 500 J

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Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

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Which of the following is not an example of kinetic energy being converted to potential energy?
KengaRu [80]

The list of choices you provided with your question
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Dmitry_Shevchenko [17]
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7 0
3 years ago
What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s
stiv31 [10]

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For us to know the kinetic energy of the vehicle,

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what is the acceleration of each body of mass 5kg rests on a frictionless table and is connected to a cable that passes over a p
Oksanka [162]

Answer:

6.53 m/s²

Explanation:

Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.

Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:

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m₂g - T = m₂a    (2)

substituting T = m₁a in equation 2:

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Both objects have an acceleration of 6.53 m/s²

8 0
3 years ago
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