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3 years ago

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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)?

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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