# 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?

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• Algol
Lv 7

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

= -1.17m/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.

• Anonymous
5 years ago

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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)

PUCK -

Vf2 = [(2m1)/(m1+m2)]Vo1

Vf2 = [(2*0.429)/(0.429 + 0.909)](3.24)

Vf2 = 2.077 ms (in the East direction)