Newton's Third Law Problem [Checking Solution]

[oMovements]

Homework Statement
Two hockey players are standing stationary on the ice facing one another. Player A has a mass of 85kg and Player B is 110kg. They shove each other and player A ends up with a velocity of 1.5m/s [W]. Find the final velocity of Player B.

Solution
F_a=M_aa_a
F_a=(85)(1.5-0/t)

F_b=M_ba_b
F_b=(110)(1.5-0/t)

Sub F_b into F_a
(110)(V_fb-0/t) = (85)(1.5-0/t)
110V_fb=127.5
V_fb=1.2 m/s [E]

 

[rock.freak667]

Your answer is correct but what you are really doing is a calculation due to conservation of linear momentum

linear momentum = mass x velocity.

so 110*Vfb = 85*1.5

 

[Andrew Mason]

F_b=M_ba_b
F_b=(110)(1.5-0/t)

I think you meant Vfb rather than 1.5 here.

Sub F_b into F_a
(110)(V_fb-0/t) = (85)(1.5-0/t)
110V_fb=127.5
V_fb=1.2 m/s [E]

What you have written is correct in the result, but you should not, and need not, assume that the acceleration is constant over the entire collision. So it would be more correct to say that:

F_{a on b} = m_bdv_b/dt = -F_{b on a} = -m_adv_a/dt

Since $v_a = \int_0^t \frac{dv_a}{dt} dt$ by integrating both sides of the above equation you see that:

m_bv_b = -m_av_a

which is just conservation of momentum, as rock.freak has pointed out.

AM