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You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sofia, with mass 70.0 kg , is given a push and slides eastward. Elena, with mass 60.0 kg , is sent sliding northward. They collide, and after the collision Sofia is moving at 38.0 ∘ north of east with a speed of 5.20 m/s and Elena is moving at 18.0 ∘ south of east with a speed of 9.00 m/s .
What was the speed of each person before the collision?
a. Elena's speed:
b. Sofia's speed:
c. By how much did the total kinetic energy of the two people decrease during the collision?

Respuesta :

Answer:

Part a)

So speed of Sofia is 11.4 m/s towards East

Part b)

Speed of Elena is 0.95 m/s

Part c)

[tex]\Delta k = 1199.3 J[/tex]

Explanation:

As we know that both are moving without friction so here total momentum is conserved before and after collision

Part a)

so we have momentum conservation along x direction given as

[tex]70 v_1 = 70(5.20 cos38) + 60(9 cos18)[/tex]

[tex]v_1 = 4.09 + 7.33[/tex]

[tex]v_1 = 11.4 m/s[/tex]

So speed of Sofia is 11.4 m/s towards East

Part b)

Similarly momentum conservation in Y direction

[tex]60 v_2 = 70 (5.20sin 38) - 60 (9sin 18)[/tex]

[tex]v_2 = 3.73 - 2.78[/tex]

[tex]v_2 = 0.95 m/s[/tex]

Speed of Elena is 0.95 m/s

Part c)

Decrease in kinetic energy of two is given as

[tex]\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)[/tex]

[tex]\Delta k = \frac{1}{2}(70)(11.4^2 - 5.20^2) + \frac{1}{2}(60)(0.95^2 - 9^2)[/tex]

[tex]\Delta k = 3602.2 - 2402.9[/tex]

[tex]\Delta k = 1199.3 J[/tex]

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