
Prove impulse- momentum theorem.
Answer
567.3k+ views
Hint:Impulse momentum theorem equals the impulse and the change in the momentum of the body. From the formula of the force, apply integrals in it with the limits of time and the momentum and apply the impulse formula to derive the impulse momentum theorem of the body.
Useful formula:
(1) The formula of force is given by
$\vec F = \dfrac{{d\vec p}}{{d\vec t}}$
Where $F$ is the force acting on the body, $\dfrac{{d\vec p}}{{d\vec t}}$ is the rate of change of momentum with respect to that of time.
(2) The formula of the impulse is given by
$J = \vec Ft$
Where $J$ is the impulse and $t$ is the time taken.
Complete step by step solution:
The impulse- momentum theorem states that the impulse is equal to the change in the momentum.
In order to prove the above theorem, the formula of the force is taken,
$\vec F = \dfrac{{d\vec p}}{{d\vec t}}$
$\vec Fd\vec t = d\vec p$
By integrating the above equation, we get
$\int {\vec Fd\vec t} = \int {d\vec p} $
Let us assume that the force is the constant and the ${p_1}$ and ${p_2}$ are the momentum of the body at the time $t = 0$ and $t = t$ respectively. Substituting these parameters in the above equation, we get
$\int\limits_0^t {\vec Fd\vec t} = \int\limits_{{p_1}}^{{p_2}} {d\vec p} $
$\vec F\int\limits_0^t {d\vec t} = \int\limits_{{p_1}}^{{p_2}} {d\vec p} $
By substituting the limit in the integral, we get
$\vec F\left( {t - 0} \right) = \left[ p \right]_{{{\vec p}_1}}^{{{\vec p}_2}}$
By simplifying the above equation, we get
\[\vec Ft = \left[ {{{\vec p}_1} - {{\vec p}_2}} \right]\]
From the formula of the impulse, it is clear that the
$J = \left[ {{{\vec p}_1} - {{\vec p}_2}} \right]$
Hence the impulse of the body is equal to the change in the momentum of it.
Hence proved.
Note:Remember that while integrating the parameters, the constant integral is constant. So it is taken outside the integral. While applying the limits, the upper limit is applied first and then from it the lower limit is subtracted. In collision of the bodies, the body experiences the impulse that is due to change in momentum of it.
Useful formula:
(1) The formula of force is given by
$\vec F = \dfrac{{d\vec p}}{{d\vec t}}$
Where $F$ is the force acting on the body, $\dfrac{{d\vec p}}{{d\vec t}}$ is the rate of change of momentum with respect to that of time.
(2) The formula of the impulse is given by
$J = \vec Ft$
Where $J$ is the impulse and $t$ is the time taken.
Complete step by step solution:
The impulse- momentum theorem states that the impulse is equal to the change in the momentum.
In order to prove the above theorem, the formula of the force is taken,
$\vec F = \dfrac{{d\vec p}}{{d\vec t}}$
$\vec Fd\vec t = d\vec p$
By integrating the above equation, we get
$\int {\vec Fd\vec t} = \int {d\vec p} $
Let us assume that the force is the constant and the ${p_1}$ and ${p_2}$ are the momentum of the body at the time $t = 0$ and $t = t$ respectively. Substituting these parameters in the above equation, we get
$\int\limits_0^t {\vec Fd\vec t} = \int\limits_{{p_1}}^{{p_2}} {d\vec p} $
$\vec F\int\limits_0^t {d\vec t} = \int\limits_{{p_1}}^{{p_2}} {d\vec p} $
By substituting the limit in the integral, we get
$\vec F\left( {t - 0} \right) = \left[ p \right]_{{{\vec p}_1}}^{{{\vec p}_2}}$
By simplifying the above equation, we get
\[\vec Ft = \left[ {{{\vec p}_1} - {{\vec p}_2}} \right]\]
From the formula of the impulse, it is clear that the
$J = \left[ {{{\vec p}_1} - {{\vec p}_2}} \right]$
Hence the impulse of the body is equal to the change in the momentum of it.
Hence proved.
Note:Remember that while integrating the parameters, the constant integral is constant. So it is taken outside the integral. While applying the limits, the upper limit is applied first and then from it the lower limit is subtracted. In collision of the bodies, the body experiences the impulse that is due to change in momentum of it.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

