91影视

The momentum of a particle: The momentum $\stackrel{鈫�}{\mathit{p}}$ of a particle is a vector quantity equal to the product of the particle鈥檚 mass mand velocity $\stackrel{\mathbf{鈫�}}{\mathbf{v}}$.

$\stackrel{\mathbf{鈫�}}{\mathbf{p}}\mathbf{=}\mathit{m}\stackrel{\mathbf{鈫�}}{\mathbf{v}}$

Newton鈥檚 second law says that the net force on a particle is equal to the rate of change of the particle鈥檚 momentum.

$\stackrel{}{\mathbf{鈭�}\mathbf{F}}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}$

Given in the question,

Force,$\stackrel{鈫�}{F}=\left(0.280N/s\right)t\hat{i}+\left(-0.450N/s\right)t^2\hat{j}$

Initial momentum${\stackrel{鈫�}{p}}_1=\left(-3.00kg.m/s\right)\hat{i}+\left(4.00kg.m/s\right)\hat{j}$

$t_1=0$and$t_2=2.00s$

From Newton鈥檚 second law

$$\begin{gathered} \underset{}{鈭�\stackrel{鈫�}{F}=\frac{d\stackrel{鈫�}{p}}{dt}} \\ d\stackrel{鈫�}{p}鈭�\stackrel{鈫�}{F}dt \\ 鈭�\stackrel{鈫�}{p}={鈭�}_{t1}^{t2}\stackrel{鈫�}{F}dt \end{gathered}$$

Substituting the values

$$\begin{gathered} 鈭�\stackrel{鈫�}{p}={鈭�}_0^2\left(0.280\right)t\hat{i}+\left(-0.450\right)t^2\hat{j}dt \\ 鈭�\stackrel{鈫�}{p}=\left(0.280\right){\left[\frac{t^2}{2}\right]}_0^{2}i+\left(-0.450\right){\left[\frac{t^3}{3}\right]}_0^2\hat{j} \\ 鈭�\stackrel{鈫�}{p}=\left(0.280\right)\left[\frac{2^2}{2}-0\right]\hat{i}+\left(-0.450\right)\left[\frac{2^2}{2}-0\right]\hat{j} \\ 鈭�\stackrel{鈫�}{p}=0.56\hat{j}-1.2\hat{j}-1.2\hat{j}kg.m/s \end{gathered}$$

We know

$$\begin{gathered} 鈭�\stackrel{鈫�}{p}=\stackrel{鈫�}{p_2}-\stackrel{鈫�}{p_1} \\ 0.56\hat{i}-1.2\hat{j}={\stackrel{鈫�}{p}}_2-\left(-3.00\right)\hat{i}+\left(4.00\right)\hat{j} \\ {\stackrel{鈫�}{p}}_2=\left(-2.44\right)\hat{i}+\left(2.80\right)\hat{j}kg.m/s \end{gathered}$$

The momentum of the box at$t=2.00s$ is $\stackrel{鈫�}{p_2}=\left(-2.44kg.m/s\right)\hat{i}+\left(2.80kg.m/s\right)\hat{j}$