# 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?

Relevance

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

My life is headed for the next minute of NOW; as, I can safely assume, yours is as well? You made it here, "Now"; however, that "Now" you will never have again. It is gone, and all you have is this New "Now", at this moment; which also has just passed. Matter of perspective. I am Retired and there is only one direction I am destined to take, and I am going too have one Hell of a Good Time getting there! In the meantime I will just have too be satisfied with Now.

• 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)