# 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
Fa=Maaa
Fa=(85)(1.5-0/t)

Fb=Mbab
Fb=(110)(1.5-0/t)

Sub Fb into Fa
(110)(Vfb-0/t) = (85)(1.5-0/t)
110Vfb=127.5
Vfb=1.2 m/s [E]

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## Answers and Replies

Homework Helper
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

Science Advisor
Homework Helper
Fb=Mbab
Fb=(110)(1.5-0/t)
I think you meant Vfb rather than 1.5 here.
Sub Fb into Fa
(110)(Vfb-0/t) = (85)(1.5-0/t)
110Vfb=127.5
Vfb=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:

Fa on b = mbdvb/dt = -Fb on a = -madva/dt

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

mbvb = -mava

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

AM