Captain America throws his shield to the right while standing still on frictionless ice, as shown below. His shield, which has a mass of 15.0 kg, leave Cap’s hand at 4.03 m/s. If Captain America has a mass of 99.8 kg, at what velocity does Captain America travel at after the throw? Include sign for direction, with to the right as positive.

When an object of mass [tex]m[/tex] travels at a velocity of [tex]{\bf v}[/tex], the momentum of that object would be [tex]{\bf p} = m\, {\bf v}[/tex].

Before throwing the shield, since velocity is zero for both the shield and the person, momentum would also be zero.

After throwing the shield:

Momentum of the [tex]15.0\; \text{kg}[/tex] shield travelling at [tex](+4.03)\; {\rm m\cdot s^{-1}}[/tex] (to the right) would be:
[tex](15.0\; \text{kg})\, (4.03\; {\rm m\cdot s^{-1}})[/tex].
Let [tex]{\bf v}[/tex] denote the current velocity of the [tex]99.8\; {\rm kg}[/tex] person. Momentum of this person would be:
[tex](99.8\; \text{kg})\, ({\bf v})[/tex].

Under the assumptions in this question, momentum would be conserved before and after throwing the shield. In other words, the sum of momentum after throwing the shield would be the same as the value before throwing the shield:

[tex](\text{total momentum, before}) = (\text{total momentum, after})[/tex].

[tex]0\; {\rm kg \cdot m\cdot s^{-1}} = (15.0\; \text{kg})\, (4.03\; {\rm m\cdot s^{-1}}) + (99.8\; \text{kg})\, ({\bf v})[/tex].

Solve this equation for [tex]{\bf v}[/tex]:

[tex]\displaystyle {\bf v} = \frac{-(15.0\; \text{kg})\, (4.03\; {\rm m\cdot s^{-1}})}{(99.8\; \text{kg})} \approx (-0.606)\; {\rm m\cdot s^{-1}}[/tex].

The sign of velocity is negative, meaning that the person would be moving opposite to the positive direction. Since motion to the right is assumed to be positive, the negative velocity means that the person would be moving to the left.