Asked by Gorg
An 80g particle moving with an initial velocity of (50m/s)î strikes and sticks to a 60g particle moving at (50m/s)j. How much kinetic energy is lost in this collision?
Answers
Answered by
Henry
M1*V1 + M2*V2 = M1*V + M2*V
0.08*50i+0.06*50j = 0.08V+0.06V.
4i + 3j = 0.14V.
5[36.9o] = 0.14V,
V = 35.7m/s[36.9o].
KE before the collision:
KE1 = 0.5M1*V1^2 + 0.5M2V2^2.
i^2 = 1, j^2 = 1.
KE after the collision:
KE2 = 0.5M1*V^2 + 0.5M2*V^2.
KE(Lost) = KE1-KE2.
0.08*50i+0.06*50j = 0.08V+0.06V.
4i + 3j = 0.14V.
5[36.9o] = 0.14V,
V = 35.7m/s[36.9o].
KE before the collision:
KE1 = 0.5M1*V1^2 + 0.5M2V2^2.
i^2 = 1, j^2 = 1.
KE after the collision:
KE2 = 0.5M1*V^2 + 0.5M2*V^2.
KE(Lost) = KE1-KE2.
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