Question 17

The two forces act at the same time, so add them to get the net force. Collect the $\hat{\boldsymbol{\jmath}}$ parts and the $\hat{\boldsymbol{k}}$ parts.

$$\begin{aligned}\boldsymbol{F}_{net}&=\boldsymbol{F}_1+\boldsymbol{F}_2\\&=\left(5t\hat{\boldsymbol{\jmath}}-3\hat{\boldsymbol{k}}\right)+\left(-t\hat{\boldsymbol{\jmath}}+\hat{\boldsymbol{k}}\right)\\&=4t\hat{\boldsymbol{\jmath}}-2\hat{\boldsymbol{k}},\quad0\le t\le2\end{aligned}$$

The component form $\begin{pmatrix}0&4t&-2\end{pmatrix}$ is also accepted. This mark can be implied by your later working.

Newton’s second law works with vectors: $\boldsymbol{F}=m\boldsymbol{a}$. Set the net force equal to $2\boldsymbol{a}$ and divide each component by the mass.

$$\begin{aligned}4t\hat{\boldsymbol{\jmath}}-2\hat{\boldsymbol{k}}&=2\boldsymbol{a}\\\boldsymbol{a}&=2t\hat{\boldsymbol{\jmath}}-\hat{\boldsymbol{k}}\end{aligned}$$

Equivalent results are accepted, such as $\boldsymbol{a}=2t\hat{\boldsymbol{\jmath}}-\hat{\boldsymbol{k}}$. If your net force was wrong, you can still earn this mark by follow-through.

Velocity is the integral of acceleration with respect to time. Integrate each component and add a constant vector $\boldsymbol{c}$.

$$\begin{aligned}\boldsymbol{v}&=\int\boldsymbol{a}\,dt\\&=\int\left(2t\hat{\boldsymbol{\jmath}}-\hat{\boldsymbol{k}}\right)dt\\&=t^2\hat{\boldsymbol{\jmath}}-t\hat{\boldsymbol{k}}+\boldsymbol{c}\end{aligned}$$

This mark can be implied by your later working.

Time is counted from the instant the forces switch on. At that instant the object still has its constant velocity, so use $\boldsymbol{v}=3\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{k}}$ at $t=0$ to find $\boldsymbol{c}$.

$$\begin{aligned}\text{At }t=0\text{:}\quad3\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{k}}&=0+\boldsymbol{c}\\\boldsymbol{c}&=3\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{k}}\end{aligned}$$

$$\begin{aligned}\boldsymbol{v}&=t^2\hat{\boldsymbol{\jmath}}-t\hat{\boldsymbol{k}}+3\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{k}}\\&=3\hat{\boldsymbol{\imath}}+t^2\hat{\boldsymbol{\jmath}}+(1-t)\hat{\boldsymbol{k}}\end{aligned}$$

Equivalent forms of this velocity are accepted. The QCAA’s second method reaches the same place one component at a time. It writes $\frac{dv_x}{dt}=0$, $\frac{dv_y}{dt}=2t$ and $\frac{dv_z}{dt}=-1$, integrates each to get $v_x=c_1$, $v_y=t^2+c_2$ and $v_z=-t+c_3$, then uses the starting velocity to find $c_1=3$, $c_2=0$ and $c_3=1$. It earns the same marks.

Momentum is mass times velocity, $m\boldsymbol{v}$. Write it in general, then substitute $t=1$.

$$m\boldsymbol{v}=2\left(3\hat{\boldsymbol{\imath}}+t^2\hat{\boldsymbol{\jmath}}+(1-t)\hat{\boldsymbol{k}}\right)$$

$$\text{At }t=1\text{:}\quad m\boldsymbol{v}=2\left(3\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}\right)$$

Equivalent forms are accepted, such as $6\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}$. This mark can be implied by your later working.

The magnitude is the length of the momentum vector. Keep the factor of $2$ outside the root, as the sample response does, and leave the answer exact.

$$\begin{aligned}\left|m\boldsymbol{v}\right|&=2\sqrt{3^2+1^2}\\&=2\sqrt{10}\ \text{kg m s}^{-1}\end{aligned}$$

Equivalent forms are accepted, such as $\sqrt{40}$.