Question

At a certain instant a particle is moving in the $\mathbf{+}\mathbf{x}$direction with momentum $\mathbf{+}\mathbf{8}\mathbf{}\raisebox{1ex}{$\mathbf{kg}\mathbf{.}\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$. During the next $\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{}\mathbf{s}$a constant force acts on the particle, with ${\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{7}\mathbf{}\mathbf{N}$and ${\mathbf{F}}_{\mathbf{y}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{}\mathbf{N}$. What is the magnitude of the momentum of the particle at the end of this$\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{}\mathbf{s}$ interval?

Short answer

The magnitude of the particle's momentum at the end of this$0.13\mathrm{s}$ interval is $8.26\raisebox{1ex}{$\mathrm{kg}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

Step-by-step solution

Identification of the given data

The given data can be listed below as

• The particle has the momentum of $+8\raisebox{1ex}{$\mathrm{kg}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

• The force acts on the particle $0.13\mathrm{s}$.

• The constant force which is acting on the particle is${\mathrm{F}}_{\mathrm{x}}=-7\mathrm{N}$and${\mathrm{F}}_{\mathrm{y}}=+5\mathrm{N}$ , respectively.

Significance of the momentum principle for the particle

This principle states that if there is a collision between two objects, the total momentum before and after the collision will be equal as there is no external force.

The equation of the principle of momentum gives the magnitude of the momentum of the particle.

Determination of the magnitude of the momentum of the particle

From the momentum principle, the magnitude of the momentum of the particle can be expressed as:

${\mathrm{p}}_2={\mathrm{p}}_1$

Hererole="math" localid="1657968279264" ${\mathrm{p}}_2$ is the final momentum of the particle and${\mathrm{p}}_1$ is the initial momentum of the particle.

${\mathrm{F}}_{\mathrm{not}}$is the net force acting on the body that can be expressed as:

role="math" localid="1657968593244" ${\mathrm{F}}_{\mathrm{net}}=\sqrt{{\mathrm{F}}_{\mathrm{x}}^2+{\mathrm{F}}_{\mathrm{y}}^2+{\mathrm{F}}_{\mathrm{z}}^2}$

For role="math" localid="1657968797682" ${\mathrm{F}}_{\mathrm{x}}=-7\mathrm{N}$,${\mathrm{F}}_{\mathrm{y}}=+5\mathrm{N}$and ${\mathrm{F}}_{\mathrm{z}}=0$net force can be calculated as:

role="math" localid="1657968861868" ${\mathrm{F}}_{\text{ret}}=\sqrt{(-7\mathrm{N}{)}^2+(+5\mathrm{N}{)}^2+(0\mathrm{N}{)}^2}$

role="math" localid="1657968881389" ${\mathrm{F}}_{\text{ret}}=\sqrt{49+25}\times \left(1\mathrm{N}\right)$

${\mathrm{F}}_{\mathrm{net}}=8.6027\mathrm{N}$

The time interval$∆\mathrm{t}=0.13\mathrm{s}$

For $∆\mathrm{t}=0.13\mathrm{s}$,${\mathrm{F}}_{\mathrm{net}}=8.6027\mathrm{N}$, and ${\mathrm{P}}_1=+8\raisebox{1ex}{$\mathrm{kg}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$; from the equation,$\left(1\right)$ the Final momentum${\mathrm{p}}_2$ can be calculated as:

${\mathrm{p}}_2=8\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}+8.6027\mathrm{N}\times 0.13\mathrm{s}$

${\mathrm{p}}_2=8\mathrm{kg}\times \mathrm{m}/\mathrm{s}+(8.6027\times 0.13)\times \left(1\mathrm{N}\times 1\mathrm{s}\times \frac{1\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)$

${\mathrm{p}}_2=8\mathrm{kg}\times \mathrm{m}/\mathrm{s}+1.118351\mathrm{kg}\times \mathrm{m}/\mathrm{s}$

${\mathrm{p}}_2=9.118351\mathrm{kg}\times \mathrm{m}/\mathrm{s}$

Thus, the magnitude of the momentum of the particle at the end of $0.13\mathrm{s}$the time interval is $9.118351\mathrm{kg}\times \mathrm{m}/\mathrm{s}$.