(B) According to Newton's second law,the rate of change of momentum is equal to the applied force:$\frac{dp}{dt} = F = kt$Integrating both sides with

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(B) According to Newton's second law,the rate of change of momentum is equal to the applied force:$\frac{dp}{dt} = F = kt$Integrating both sides with respect to time from $t = 0$ to $t = T$ and momentum from $p$ to $3p$:$\int_{p}^{3p} dp = \int_{0}^{T} kt \, dt$$[p]_{p}^{3p} = k \left[ \frac{t^2}{2} \right]_{0}^{T}$$3p - p = k \left( \frac{T^2}{2} - 0 \right)$$2p = \frac{kT^2}{2}$$4p = kT^2$$T^2 = \frac{4p}{k}$$T = 2\sqrt{\frac{p}{k}}$

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

Medium

$A$ particle of mass $m$ is moving in a straight line with momentum $p$. Starting at time $t = 0$,a force $F = kt$ acts in the same direction on the moving particle during time interval $T$ so that its momentum changes from $p$ to $3p$. Here $k$ is a constant. The value of $T$ is

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A
$2\sqrt{\frac{k}{p}}$
B
$2\sqrt{\frac{p}{k}}$
C
$\sqrt{\frac{2k}{p}}$
D
$\sqrt{\frac{2p}{k}}$

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Class 11

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Newton's Laws of Motion and Friction

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Second Law of Motion

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$A$ particle of mass $m$ is moving in a straight line with momentum $p$. Starting at time $t = 0$,a force $F = kt$ acts in the same direction on the moving particle during time interval $T$ so that its momentum changes from $p$ to $3p$. Here $k$ is a constant. The value of $T$ is

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The maximum acceleration of a $5\,kg$ block is $.............\,m/s^{2}$.

Write the symbolic form of Newton's second law.

The force required to stop a body of mass $10 \,kg$ moving along a straight line path with a velocity of $10 \,ms^{-1}$ in a time of $10 \,s$ is (in $\,N$)

$A$ block of mass $1 \ kg$ starts from rest at $x=0$ and moves along the $X$-axis under the action of a force $F=kt$, where $t$ is time and $k=1 \ Ns^{-1}$. The distance the block will travel in $6 \ seconds$ is: (in $m$)

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