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AysviL [449]
3 years ago
6

Starting with an initial speed of 6.63 m/s at a height of 0.473 m, a 1.67-kg ball swings downward and strikes a 5.91-kg ball tha

t is at rest, as the drawing shows.
(a) Using the principle of conservation of mechanical energy, find the speed of the 1.67-kg ball just before impact.
(b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 1.67-kg ball just after the collision.
(c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.91-kg ball just after the collision.
(d) How high does the 1.67-kg ball swing after the collision, ignoring air resistance?
(e) How high does the 5.91-kg ball swing after the collision, ignoring air resistance?
Physics
1 answer:
Jet001 [13]3 years ago
5 0

Answer:

know the truth.’ Then I said: ‘Blessed be God, who put this in the Chan’s heart. But our Scriptures tell us, the servant of God should not dispute, but should show mildness

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R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

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R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

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R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

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