Question

a toy train starts from rest and is drive along a horizontal track by a motor which exerts a constant driving force the effects of friction and air resistance can be neglected which of the following graphs best represents the momentum M of train as a function of distance

Answer 1

Given:

A toy train starts from rest and is drive along a horizontal track by a motor which exerts a constant driving force.

To find:

Graph between Momentum and distance.

Solution:

We know that force is defined as the instantaneous rate of change of momentum:

$\sf{ \therefore \: F = \dfrac{dM}{dt}}$

Now , the question says that the force is constant:

$\sf{ \therefore \: F = \dfrac{dM}{dt} = k \: \: \: \: \: ....(let)}$

$\sf{ = > \: dM = k dt}$

$\displaystyle \sf{ = > \: \int dM = k \int \: dt}$

$\displaystyle \sf{ = > \: M = kt \: \: \: .......(1)}$

Again , we can say that acceleration of train is constant :

$\sf{\therefore \: a = constant = c}$

$\sf{ = > \: \dfrac{dv}{dt} = c}$

$\displaystyle \sf{ = > \: \int dv= c \: \int dt}$

$\displaystyle \sf{ = > \: v= c t}$

$\displaystyle \sf{ = > \: \dfrac{dx}{dt} = c t}$

$\displaystyle \sf{ = > \: \int dx = c \: \int t \: dt}$

$\displaystyle \sf{ = > \: x = \dfrac{c {t}^{2} }{2} }$

$\displaystyle \sf{ = > \: t = \sqrt{ \frac{2x}{c} } \: \: \: \: \: .......(2) }$

Putting value of c in eq (1):

$\displaystyle \sf{ = > \: M = kt \: \: \: .......(1)}$

$\displaystyle \sf{ = > \: M = k \times \sqrt{ \frac{2x}{c} } }$

$\displaystyle \sf{ = > \: M \propto \: {x}^{ \frac{1}{2} } }$

So, the graph will be as attached.

HOPE IT HELPS.