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$\begin{array}{l}{\text { At time } t=0, \text { a } 2150-\mathrm{kg} \text { rocket in outer space fires }} \\ {\text { an engine that exerts an increasing force on it in the }+x \text { -direction. }} \\ {\text { This force obeys the equation } F_{x}=A t^{2}, \text { where } t \text { is time, and has a }} \\ {\text { magnitude of } 781.25 \mathrm{N} \text { when } t=1.25 \mathrm{s} \text { . (a) Find the SI value of }} \\ {\text { the constant } A, \text { including its units. (b) What impulse does the }} \\ {\text { engine exert on the rocket during the } 1.50 \text { -s interval starting } 2.00 \text { s }} \\ {\text { after the engine is fired? (c) By how much does the rocket's veloc- }} \\ {\text { ity change during this interval? }}\end{array}$

$F_{x}=A t^{2}$

$m=2150 \mathrm{kg}$

${J=\int_{t_1}^{t_2} \vec{F} d t \longrightarrow J_{x}=\int_{t_1}^{t_ 2} F_{x} d t}$

$F_{x}=781.25 \mathrm{N} \quad@t=1.25 \mathrm{s}$

$\\ {F_{x}=A t^{2} \longrightarrow A=\frac{F_ x}{t^{2}}=\frac{781.25}{(1.25)^{2}}=500 \mathrm{N} / \mathrm{s}^{2}}$

$J_{x}=\int_{2}^{3.5} At^2 d t=\frac{1}{3} A {t_{2}}^{3}-\frac{1}{3} A {t_{1}}^{3}=\frac{1}{3} A\left[{t_{2}}^{3}-{t_{1}}^{3}\right]$

$J _x=\frac{1}{3}(500)\left[3.5^{3}-2^{3}\right]=5.81 \times 10^{3} \quad N .S$

$J _x=P_{x_{2}}-P_{x_ 1}=m v_{x_{2}}-m v_{x_1}=m\left(v_{x_ 2}-v_{x_{1}}\right)$

$∴ \quad J_ x=m \Delta V _x \rightarrow \Delta V_{x}=\frac{J_ x}{m}=\frac{5.81 \times 10^{3}}{2150}=2.7 \mathrm{m/}{s}$

$\begin{array}{l}{\text { A } 2.00 \text { -kg stone is sliding }} \\ {\text { to the right on a frictionless hori- }} \\ {\text { zontal surface at } 5.00 \mathrm{m} / \mathrm{s} \text { when }} \\ {\text { it is suddenly struck by an object }} \\ {\text { that exerts a large horizontal }} \\ {\text { that exerts a large horizontal }} \\ {\text { force on it for a short period of }} \\ {\text { shows the magnitude of this force }} \\ {\text { stops acting, find the magnitude }} \\ {\text { and direction of the stone's veloc }} \\ {\text { if the force acts (i) to the right }} \\ {\text { or (ii) to the left. }}\end{array}$

$m=2 k g \quad, \quad v_{1}=5 m/ s \longrightarrow$

$J=\int F d t= \text { Area under the Curve }$

$J=\left(2.5 * 10^{3}\right)\left(1 * 10^{-3}\right)=2.5 \mathrm{N} . \mathrm{s}$

$v_{2}=? ? \rightarrow \text { to the right } \rightarrow J_ x \ (+)$

$J_ x=m\left(v_{2}-v_{1}\right)=v_{2} - v_{1}=\frac{J_{x}}{m} \rightarrow v_{2}=\frac{J_{x}}{m}+v_{1}$

$∴ V_{2}=\frac{2.5}{2}+5=6.25 \mathrm{m} / \mathrm{s}$

$V_{2}=\frac{J_ x}{m}+V_{1}$

$v_{2} \stackrel{\text { ?? }}{\rightarrow} \text { to the left } \rightarrow J_ x \ (-)$

$∴ V_{2}=\frac{-2.5}{2}+5=3.75 \mathrm{m/s}$

$\begin{array}{l}{\text { Starting at } t=0, \text { a horizontal net force } \overrightarrow{\boldsymbol{F}}=} \\ {(0.280 \mathrm{N} / \mathrm{s}) t\left(-0.450 \mathrm{N} / \mathrm{s}^{2}\right) t^{2} \hat{\boldsymbol{j}} \text { is applied to a box that }} \\ {\text { has an initial momentum } \vec{p}=(-3.00 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\imath}+(4.00 \mathrm{kg} \text { . }} \\ {\mathrm{m} / \mathrm{s}) \hat{\boldsymbol{j}} . \text { What is the momentum of the box at } t=2.00 \mathrm{s} ?}\end{array}$

$t=0$

$\vec{F}=0.28 t \ i+ (-0.45) t^{2} \hat J \quad \ P_{2} \ ?? @ t=2s$

$\vec{P}_{1}=-3 \hat{i}+{4} \hat{j}$

(x)

$F_{x}=0.28 t \quad \mathrm{N} / \mathrm{s}$

$P_{x}=-3 \quad k g \cdot m / s$

$J_{x}=\int_{t_{1}}^{t_{2}} F_{x} d t=\int_{0}^{2} 0.28 t \ d t$

$=0.28\left[\frac{1}{2}\left(2 )^{2}-0^{2}\right]=0.56 \mathrm{N} \cdot \mathrm{S}\right.$

$P_{2 x}=P_{1 x}+J_{x} \Rightarrow-3+0.56=-2.44 kg. m/s$

(y)

$F_ y=-0.45 t^{2}$

$P_{y}=4 \quad k g . m / s$

$J_{y}=\int_{t_1}^{t_{2}} F_{y} d t=\int_{0}^{2}\left(-0.45 t^{2}\right) d t=$

$J_{y}=(-0.45)\left(\frac{1}{3}\right)\left[\left(2^{2}\right)-(0)^{2}\right]=-1.2 N .S$

$P_{2 y}=P_{1 y}+J_ y=4-t2=2.8 kg. m/s$

$\vec{P}_{2}=-2.44 \hat i+2.8\hat j \quad k g . m / s$