A.2 Forces and momentum

Let the initial direction of travel be positive. So $v_i=+12{\text{m s}}^{-1}$ and $v_f=-8.0{\text{m s}}^{-1}$. Mass $m=0.16\text{kg}$.

$p_i=m{v}_i=0.16\times 12=+1.92{\text{kg m s}}^{-1}$

$p_f=m{v}_f=0.16\times (-8.0)=-1.28{\text{kg m s}}^{-1}$

$\Delta p=p_f-p_i=-1.28-1.92=-3.20{\text{kg m s}}^{-1}$The magnitude is $3.2{\text{kg m s}}^{-1}$.

We need the magnitude of the change in momentum for a puck that reverses direction. The key is to account for the direction reversal by assigning opposite signs to the velocities.

This comes from $0.16\times (12-8)=0.64$. This ignores the direction reversal and merely takes the difference in speeds — incorrect.

This is approximately the initial momentum alone ($0.16\times 12=1.92$), not the change in momentum — incorrect.

This equals $0.16\times 8.0=1.28$, which is just the final momentum magnitude — not the change in momentum.

With correct signs: $\Delta p=0.16(-8.0)-0.16(+12)=-1.28-1.92=-3.20{\text{kg m s}}^{-1}$, giving a magnitude of $3.2{\text{kg m s}}^{-1}$.

The question asks for the form of Newton's second law that remains valid when mass is not constant. This rules out expressions that assume constant mass.

$F_{\text{net}}=ma$ assumes constant mass because it is derived from $\frac{\Delta (mv)}{\Delta t}=m\frac{\Delta v}{\Delta t}$ only when $m$ is constant. It is a special case, not the general law.

The general statement is $F_{\text{net}}=\frac{\Delta p}{\Delta t}$, the rate of change of momentum. This holds for any situation, including rockets with variable mass.

$F_{\text{net}}=\frac{\Delta p}{\Delta t}$ is the universally correct form. When mass is constant, this reduces to $F=ma$, confirming it is the more fundamental expression.

We need the form of Newton's second law that is valid even for variable mass systems.

This is only valid when mass is constant. It is a special case derived from the momentum form, so it cannot apply to variable mass situations.

This expression is simply acceleration — it has no force units on the right side and ignores mass entirely. It is dimensionally incorrect for force.

$mv$ is momentum itself, not the rate of change of momentum. This has units of ${\text{kg m s}}^{-1}$, not $\text{N}$.

$F_{\text{net}}=\frac{\Delta p}{\Delta t}$ is the general and more fundamental statement. It equals $ma$ only when mass is constant, confirming it is the correct answer.