(C) Given: Mass $m = 5 \ kg$,initial velocity $\vec{u} = (30 \hat{i} + 40 \hat{j}) \ m/s$,and force $\vec{F} = -(\hat{i} + 5 \hat{j}) \ N$.Using Newto

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(C) Given: Mass $m = 5 \ kg$,initial velocity $\vec{u} = (30 \hat{i} + 40 \hat{j}) \ m/s$,and force $\vec{F} = -(\hat{i} + 5 \hat{j}) \ N$.Using Newton's second law,the acceleration $\vec{a}$ is given by $\vec{a} = \frac{\vec{F}}{m} = \frac{-(\hat{i} + 5 \hat{j})}{5} = (-0.2 \hat{i} - 1 \hat{j}) \ m/s^2$.The velocity at any time $t$ is given by $\vec{v} = \vec{u} + \vec{a}t$.Substituting the components,the $y$-component of velocity is $v_y = u_y + a_y t$.Here,$u_y = 40 \ m/s$ and $a_y = -1 \ m/s^2$.We want the time $t$ when $v_y = 0$.$0 = 40 + (-1)t$.$t = 40 \ s$.

Class 11 Physics Newton's Laws of Motion and Friction Second Law of Motion

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$A$ body of mass $5 \ kg$ starts from the origin with an initial velocity $(30 \hat{i} + 40 \hat{j}) \ m/s$. If a constant force $-(\hat{i} + 5 \hat{j}) \ N$ acts on the body,then the time in which the $y$-component of its velocity becomes zero is (in $s$)

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Newton's Laws of Motion and Friction

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Second Law of Motion

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The time taken by a force of $2 \ N$ to produce a change of momentum of $0.4 \ kg \ m \ s^{-1}$ in a body is: (in $s$)

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An object with mass $5 \ kg$ is acted upon by a force,$\vec{F}=(-3 \hat{i}+4 \hat{j}) \ N$. If its initial velocity at $t=0$ is $\vec{v}=(6 \hat{i}-12 \hat{j}) \ ms^{-1}$,the time at which it will just have a velocity along the $y$-axis is: (in $s$)

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State and explain Newton's second law of motion. Write its important points.

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$5\, kg$ દળ ધરાવતા એક પદાર્થ પર અચળ બળ $\overrightarrow F = {F_x}\hat i + {F_y}\hat j$ લાગે છે. $t = 0\, s$ સમયે તેનો વેગ $\overrightarrow v = (6\hat i - 2\hat j)\, m/s$ છે અને $t = 10\, s$ સમયે તેનો વેગ $\overrightarrow v = 6\hat j\, m/s$ છે. તો બળ $\overrightarrow F$ શોધો.

MediumJEE MAIN 2014

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$A$ body of mass $5 \ kg$ starts from the origin with an initial velocity $(30 \hat{i} + 40 \hat{j}) \ m/s$. If a constant force $-(\hat{i} + 5 \hat{j}) \ N$ acts on the body,then the time in which the $y$-component of its velocity becomes zero is (in $s$)

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$A$ car of mass $100 \, kg$ is moving with a velocity of $5 \, m/s$. The force required to stop it in $\frac{1}{10} \, s$ is . . . . . . $N$.

The time taken by a force of $2 \ N$ to produce a change of momentum of $0.4 \ kg \ m \ s^{-1}$ in a body is: (in $s$)

An object with mass $5 \ kg$ is acted upon by a force,$\vec{F}=(-3 \hat{i}+4 \hat{j}) \ N$. If its initial velocity at $t=0$ is $\vec{v}=(6 \hat{i}-12 \hat{j}) \ ms^{-1}$,the time at which it will just have a velocity along the $y$-axis is: (in $s$)

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