Question: The linear momentum \[p\] of a body moving in one dimension varies with time \[t\]according to the e...

Question

The linear momentum $p$ of a body moving in one dimension varies with time $t$ according to the equation $p = a + b{t^2}$ , where $a$ and $b$ are positive constants. The net force acting on the body is
A. A constant
B. Proportional to ${t^2}$
C. Inversely proportional to $t$
D. Proportional to $t$

Answer 1

Solution

Force is rate of change of momentum or it can be expressed as $F = \dfrac{{dp}}{{dt}}$ . Place the given value of $p$ in the equation.

Stepwise solution:

Given, $p = a + b{t^2}$ …… (i)
We know that, rate of change of momentum can be expressed as force.
i.e. $F = \dfrac{{dp}}{{dt}}$

Now, substitute the value of equation (i) in the above equation and solve

F = \dfrac{d}{{dt}}\left( {a + b{t^2}} \right) \\\ = b\dfrac{d}{{dt}}\left( {{t^2}} \right) \\\ = 2bt \\\

From this derivation we can say that $F \propto t$
Therefore, the net force acting on the body is proportional to $t$

Hence, option D is correct.

Additional information:

Force: A force is a kind of push or pull on an object that results from the contact of the object with some other object. There is indeed a force on each of the objects whenever there is an interaction with two objects. The two subjects do not feel the force anymore as the contact ends.

The rate of momentum shift of an object is proportional to the force applied as a factor and is now in the direction of the resulting force. The resulting force is proportional to the amount of momentum shift.

The rate of momentum shift of a body is equivalent to the result of an object's mass and acceleration produced in it regardless of the force acting on a body, assuming that the object's mass remains constant. It is presented as:
Rate of change of momentum $=$ mass $\times$ acceleration $=$ force.

Note:
In this question we are asked to determine the relation between the force and time $t$ . Force can be expressed as rate of change of momentum $\left( p \right)$ .i.e. $F = \dfrac{{dp}}{{dt}}$ . Place the given value of $p$ in the equation. Remember that derivative of a constant term is $0$ .