Can someone help me with this physics problem?

Question

Two ice skaters are holding hands at the center of a frozen pond when an argument ensues. Skater A shoves skater B along a horizontal direction.

(a) Identify the horizontal forces acting on A.
__skater B's momentum
__skater A's inertia
__gravity
__force from skater B
__normal force of ice



(b) Identify the horizontal forces acting on B.
__force from skater A
__normal force of ice
__gravity
__skater B's inertia
__skater A's momentum


(c) Which force is greater, the force on A or the force on B?
__the force on A
__the force on B
__the forces are the same size
__we need to know the skaters' masses

(d) Can conservation of momentum be used for the system of A and B?
__yes
__no

Defend your answer.


(e) If A has a mass of 0.620 times that of B, and B begins to move away with a speed of 2.25 m/s, find the speed of A
in m/s

Answer 1

a) Force from skater B
-Normal force and gravity act in the vertical direction
-Inertia is another word for mass; it's not a force
-Momentum is not force

b) Force from skater A
Same reasons as in part (a)

c) The forces are the same size
-It's Newton's 3rd law: for every action (the force from skater B on skater A) there is an equal and opposite reaction (the force from skater A on skater B)

d) Yes
Assuming the ice a frictionless surface, the skaters make up a closed system. This means that their net momentum after the push will be equal to their net momentum before the push (that is, 0 net momentum, with respect to the ice).

e) Σp,i = Σp,f
0 = mA(vA,f) + mB(vB,f)
Now solve for vA,f:
vA,f = (mB(vB,f)) / mA = (mB(vB,f)) / (0.62(mB))
= vB,f / 0.62 = 2.25 / 0.62 = 3.63 m/s

Answer 2

Hi, I try not to do homework, just help with it:

a, b, and c, all involve Newton's law of motion which states that for each and every force there is an equal and opposite force. The event was A pushing B, which occurred in the horizontal direction.

For d, A and B form a closed system, there are no outside forces. Conservation of momentum means that the velocity center of mass for a system will not change unless there is an outside force on the system.

e. The initial and final momentums for a closed system must remain the same. This means that
MaVa1 +MbVb1= MaVa2 +MbVb2

Here, we know the initial Velocities for Va1 and Vb1 are both zero, therefore

0 = MaVa2 + MbVb2

you know Vb and Ma/Mb so plug'nchug.

Hope this is useful.