A 0.429 kg ice puck, moving east with a speed of 3.24 m/s, has a head-on collision with a 0.909 kg puck initia?
speed of the ice puck?
speed of the puck?
- AlgolLv 71 decade agoFavorite Answer
The problem is apparently saying that the 0.909kg puck is initially at rest, then the law of conservation of momentum gives:
m₁v₁(i) + m₂v₂(i) = m₁v₁(f) + m₂v₂(f)
(0.429kg)(3.24m/s) + 0 = (0.429kg)v₁(f) + (0.909kg)v₂(f)
1.39kg∙m/s = (0.429kg)v₁(f) + (0.909kg)v₂(f)------------>(1)
You have two unknowns, so you need another equation. The equation which relates the velocity of approach to the velocity of recession is:
v₁(i) - v₂(i) = -[v₁(f) - v₂(f)]
3.24m/s - 0 = v₂(f) - v₁(f)
v₁(f) = v₂(f) - 3.24m/s------------>(2)
Plugging (2) into (1)
1.39kg∙m/s = (0.429kg)[v₂(f) - 3.24m/s] + (0.909kg)v₂(f)
1.39kg∙m/s = (0.429kg)v₂(f) - 1.39kg∙m/s + (0.909kg)v₂(f)
(1.34kg)v₂(f) = 2.78kg∙m/s
v₂(f) = 2.074m/s
v₁(f) can be foiund directly from (2):
v₁(f) = 2.074m/s - 3.24m/s
Note: from rounding the numbers, these answers may vary slightly from your book, but they should be close (unless I made an error in entering numbers into my calculator or something).
Hope this helps.
- Anonymous5 years ago
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- 1 decade ago
ICE PUCK -
Vf1 = [(m1-m2)/(m1+m2)]Vo1
Vf1 = [(0.429 - 0.909)/(0.429 + 0.909)](3.24)
Vf1 = -1.16 (so 1.16 m/s in the West direction)
Vf2 = [(2m1)/(m1+m2)]Vo1
Vf2 = [(2*0.429)/(0.429 + 0.909)](3.24)
Vf2 = 2.077 ms (in the East direction)