A 0.160-kg hockey puck is moving on an icy, frictionless,

horizontal surface. At $\mathit{t}\mathbf{=}\mathbf{0}$, the puck is moving to the right at $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$.

(a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 N directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 N directed to the left is applied from t= 0 to t= 0.050 s, what is the final velocity of the puck?

The momentum of a particle: The momentum $\overrightarrow{\mathbf{P}}$of the particle’s mass mand velocity $\overrightarrow{\mathbf{v}}$.

$\overrightarrow{\mathbf{p}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{v}}$

Newton’s second law

The net force on a particle is equal to the rate of change of the particle’s momentum.

$\mathbf{\sum }\mathit{F}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}$

Given in the question.

Mass of the puck, $m=0.160\mathrm{kg}$

Considering the right direction as the positive direction.

The initial velocity of the puck, $v_i=3.00\mathrm{m}/\mathrm{s}$

$$\begin{gathered} \sum F=25\mathrm{N} \\ ∆t=0.05\mathrm{s} \end{gathered}$$

According to the Newton’s second law

localid="1664431426577" $$\begin{gathered} \sum F=\frac{dp}{dt} \\ \sum F=\frac{p_2-p_1}{∆t} \\ \sum F=\frac{m{v}_f-m{v}_i}{∆t} \end{gathered}$$

Substituting the values into the formula.

$$\begin{gathered} 25=\frac{\left(0.16\right)v_f-\left(0.16\right)\left(3\right)}{0.05} \\ 0.{16}_{v_f}=25\times 0.05+0.48 \\ {v}_f=\frac{1.73}{0.16} \\ {v}_f=10.81\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence the velocity of the puck when 25N force is applied is 10.81 m/s in the right direction.

Given in the question.

Mass of the puck, $m=0.160\mathrm{kg}$

Considering the right direction as the positive direction.

The initial velocity of the puck, $v_i=3.00\mathrm{m}/\mathrm{s}$

$\sum F=-12\mathrm{N}$

$∆t=0.05\mathrm{s}$

According to the Newton’s second law

$$\begin{gathered} \sum F=\frac{dp}{dt} \\ \sum F=\frac{p_2-p_1}{∆t} \\ \sum F=\frac{m{v}_f-m{v}_i}{∆t} \end{gathered}$$

Substituting the values into the formula.

$$\begin{gathered} -12=\frac{\left(0.16\right)v_f-\left(0.16\right)\left(3\right)}{0.05} \\ 0.16v_f=-12\times 0.05+0.48 \\ {v}_f=\frac{-0.12}{0.16} \\ {v}_f=-0.75\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence the velocity of the puck when 12N force is applied is 0.75 m/s in the left direction.