A 0.500 kg sphere moving with a velocity (2.00i - 3.50j + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with a velocity (-1.00i + 2.00j - 2.80k) m/s.
(a) If the velocity of the 0.500 kg sphere after the collision is (-1.00i + 3.00j - 8.00k) m/s, find the velocity of the 1.50 kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic).
(0 i+__ j +___ k)m/s
This is a perfectly inelastic collision.
(b) If the velocity of the 0.500 kg sphere after the collision is (-0.250i + 0.625j - 1.85k) m/s, find the final velocity of the 1.50 kg sphere and identify the kind of collision.
(__ i+__ j +___ k)m/s
This is a perfectly inelastic collision.
(c) What if? If the velocity of the 0.500 kg sphere after the collision is (-1.00i + 2.50j + a k) m/s, find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a negative value and a positive value. Report each with their corresponding final velocities.)
a (positive value) __________m/s^2 v2f = ________ k m/s
a (negative value) __________ m/s^ v2f = ________ k m/s
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To solve part (c), we can use the conservation of momentum and kinetic energy to find the value of a and the velocities of the spheres after an elastic collision.
Given:
Mass of the first sphere, m1 = 0.500 kg
Initial velocity of the first sphere, u1 = (2.00i - 3.50j + 1.00k) m/s
Mass of the second sphere, m2 = 1.50 kg
Initial velocity of the second sphere, u2 = (-1.00i + 2.00j - 2.80k) m/s
Final velocity of the first sphere, v1 = (-1.00i + 2.50j + a k) m/s (where a is unknown)
Final velocity of the second sphere, v2 = v2f k m/s (where v2f is unknown)
Conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
0.500(2.00i - 3.50j + 1.00k) + 1.50(-1.00i + 2.00j - 2.80k) = 0.500(-1.00i + 2.50j + a k) + 1.50v2f
Expanding and collecting i, j, and k terms separately:
(1.00i - 1.75j + 0.50k) + (-1.50i + 3.00j - 4.20k) = (-0.50i + 1.25j + 0.50ak) + 1.50v2f
Simplifying and equating i, j, and k terms separately:
0.50i + 1.25i - 0.50i = 0
-1.75j + 3.00j - 1.25j = 0
0.50k - 4.20k - 0.50ak = 0
Solving these equations gives us:
a = 4.20
So, the positive value of a is 4.20 m/s^2, and the velocity of the second sphere after an elastic collision is:
v2f = 0.500(2.00i - 3.50j + 1.00k) + 1.50(-1.00i + 2.00j - 2.80k) - 0.500(-1.00i + 2.50j + 4.20k)
= (1.00i - 1.75j + 0.50k) + (-1.50i + 3.00j - 4.20k) + (0.50i - 1.25j - 2.10k)
= -0.50i + 0.00j - 6.05k
Therefore, the positive value of a is 4.20 m/s^2, and the velocity of the second sphere after an elastic collision is (-0.50i + 0.00j - 6.05k) m/s.
To find the negative value of a and the corresponding velocity of the second sphere after an elastic collision, we can substitute -a for a in the equation above.
So, the negative value of a is -4.20 m/s^2, and the velocity of the second sphere after an elastic collision is:
v2f = -0.50i + 0.00j + 6.05k m/s.