
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?
Answer
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Hint:
To solve this question, we need to use the mathematical form of the basic definition of the force. Then, that expression is to be converted into a definite integral. Using the boundary conditions as the limits of the definite integration, we will get the final answer.
Formula used: The formula used in solving this question is
$F = \dfrac{{dp}}{{dt}}$, where $F$ is the force, $p$ is the momentum, and $t$ is the time
Complete step by step answer:
We know that the force acting on a body is equal to the rate of the change of its momentum. Writing this definition of the force mathematically, we have
$F = \dfrac{{dp}}{{dt}}$
This can also be written as
$dp = Fdt$
Integrating both the sides, we get
$\int\limits_{{p_1}}^{{p_2}} {dp} = \int\limits_{{t_1}}^{{t_2}} {Fdt} $
According to the question, $F = kt$
$\int\limits_{{p_1}}^{{p_2}} {dp} = \int\limits_{{t_1}}^{{t_2}} {ktdt} $
Since $k$ is a constant, so it can be taken out of the integral’
$\int\limits_{{p_1}}^{{p_2}} {dp} = k\int\limits_{{t_1}}^{{t_2}} {tdt} $
$\left[ p \right]_{{p_1}}^{{p_2}} = k\left[ {\dfrac{{{t^2}}}{2}} \right]_{{t_1}}^{{t_2}}$
Substituting the limits, we get
${p_2} - {p_1} = k\left( {\dfrac{{{t_2}^2 - {t_1}^2}}{2}} \right)$
According to the question, ${p_1} = p$,${p_2} = 3p$,${t_1} = 0$, and ${t_2} = T$
Substituting these in the above equation, we get
$\therefore 3p - p = k\left( {\dfrac{{{T^2} - {0^2}}}{2}} \right)$
$2p = \dfrac{{k{T^2}}}{2}$
Multiplying by$\dfrac{2}{k}$ on both the sides, we get
${T^2} = \dfrac{{4p}}{k}$
Taking square root
$T = \sqrt {\dfrac{{4p}}{k}} $
Or, $T = 2\sqrt {\dfrac{p}{k}} $
Hence, the value of $T$ is equal to $2\sqrt {\dfrac{p}{k}} $.
Note:
These types of problems, involving the variable force, are always solved with the help of the mathematical form of the basic definition of the force. All of these problems involve the use of integration.
Do not confuse the value of the mass given in the problem. It is just the extra information, which is not related, in any way to the solution. Do not divert your approach to solve this question by trying to incorporate the value of the mass.
To solve this question, we need to use the mathematical form of the basic definition of the force. Then, that expression is to be converted into a definite integral. Using the boundary conditions as the limits of the definite integration, we will get the final answer.
Formula used: The formula used in solving this question is
$F = \dfrac{{dp}}{{dt}}$, where $F$ is the force, $p$ is the momentum, and $t$ is the time
Complete step by step answer:
We know that the force acting on a body is equal to the rate of the change of its momentum. Writing this definition of the force mathematically, we have
$F = \dfrac{{dp}}{{dt}}$
This can also be written as
$dp = Fdt$
Integrating both the sides, we get
$\int\limits_{{p_1}}^{{p_2}} {dp} = \int\limits_{{t_1}}^{{t_2}} {Fdt} $
According to the question, $F = kt$
$\int\limits_{{p_1}}^{{p_2}} {dp} = \int\limits_{{t_1}}^{{t_2}} {ktdt} $
Since $k$ is a constant, so it can be taken out of the integral’
$\int\limits_{{p_1}}^{{p_2}} {dp} = k\int\limits_{{t_1}}^{{t_2}} {tdt} $
$\left[ p \right]_{{p_1}}^{{p_2}} = k\left[ {\dfrac{{{t^2}}}{2}} \right]_{{t_1}}^{{t_2}}$
Substituting the limits, we get
${p_2} - {p_1} = k\left( {\dfrac{{{t_2}^2 - {t_1}^2}}{2}} \right)$
According to the question, ${p_1} = p$,${p_2} = 3p$,${t_1} = 0$, and ${t_2} = T$
Substituting these in the above equation, we get
$\therefore 3p - p = k\left( {\dfrac{{{T^2} - {0^2}}}{2}} \right)$
$2p = \dfrac{{k{T^2}}}{2}$
Multiplying by$\dfrac{2}{k}$ on both the sides, we get
${T^2} = \dfrac{{4p}}{k}$
Taking square root
$T = \sqrt {\dfrac{{4p}}{k}} $
Or, $T = 2\sqrt {\dfrac{p}{k}} $
Hence, the value of $T$ is equal to $2\sqrt {\dfrac{p}{k}} $.
Note:
These types of problems, involving the variable force, are always solved with the help of the mathematical form of the basic definition of the force. All of these problems involve the use of integration.
Do not confuse the value of the mass given in the problem. It is just the extra information, which is not related, in any way to the solution. Do not divert your approach to solve this question by trying to incorporate the value of the mass.
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