Start Free Mock LoginAlready registered? Login here

Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.MiscellaneousMedium2 minQB

PhysicsMediumsingle choice

An artillery piece of mass $M_{1}$ fires a shell of mass $M_{2}$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :

Options:

Answer:

Solution:

As linear momentum will be conserved, $|{\overset{⇀}{p}}_{1}| = |{\overset{⇀}{p}}_{2}|$

Now, kinetic energy can be written as $K E = \frac{p^{2}}{2 M}$ .

Also $p$ is the same for both masses. We can write,

$K E \propto \frac{1}{m}$

$\Rightarrow \frac{K E_{1}}{K E_{2}} = \frac{\frac{p^{2}}{2 M_{1}}}{\frac{p^{2}}{2 M_{2}}} = \frac{M_{2}}{M_{1}}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...