(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

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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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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

Similar Questions

$A$ $4 \ kg$ object has a velocity of $3.0 \hat{i} \ m/s$ at some instant. Eight seconds later,its velocity is $(8.0 \hat{i} + 10.0 \hat{j}) \ m/s$. Assuming that the object is subjected to a constant net force,the magnitude of the force is

$A$ bullet of mass $20 \,g$ moving with $500 \,ms^{-1}$ is pierced $1 \,cm$ into a wooden block. The retarding force experienced by the bullet is:

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