The linear momentum p of a body moving in one | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Newton's second law of motion,the net force $F$ acting on a body is equal to the rate of change of its linear momentum $p$ with respe

Vedclass · Accepted Answer

Proportional to $t$

Vedclass · Answer

(D) According to Newton's second law of motion,the net force $F$ acting on a body is equal to the rate of change of its linear momentum $p$ with respect to time $t$. Mathematically,$F = \frac{dp}{dt}$. Given the equation for momentum is $p = a + bt^2$. Differentiating this equation with respect to time $t$,we get: $F = \frac{d}{dt}(a + bt^2)$. Since $a$ is a constant,its derivative is $0$. $F = 0 + b(2t) = 2bt$. Since $2$ and $b$ are constants,we can conclude that $F \propto t$. Therefore,the net force acting on the body is proportional to $t$.

The linear momentum $p$ of a body moving in one dimension varies with time according to the equation $p = a + bt^2$,where $a$ and $b$ are positive constants. The net force acting on the body is

Similar Questions

Which physical quantity is represented by the first derivative of linear momentum with respect to time?

$A$ constant force $F$ acts on two different masses $m_1$ and $m_2$ independently and produces accelerations $A_1$ and $A_2$ respectively. When the same force $F$ acts on their combined mass $(m_1 + m_2)$,the acceleration $A$ produced is:

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 $\,N$)

$A$ body of mass $2 \, kg$ is moving with a velocity $8 \, m/s$ on a smooth surface. If it is to be brought to rest in $4 \, seconds$,then the force to be applied is ......... $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module