Kinematics Question 262

Question: If a particle of mass m is moving with constant velocity v parallel to x-axis in x-y plane as shown in fig. it’s angular momentum with respect to origin at any time t will be

Options:

A) $ mvb\hat{k} $

B) $ -mvb\hat{k} $

C) $ mvb\hat{i} $

D) $ mv\hat{i} $

Show Answer

Answer:

Correct Answer: B

Solution:

We know that, Angular momentum $ \overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p} $ in terms of component becomes $ \overrightarrow{L}= \begin{vmatrix} \hat{i} & \hat{j} & {\hat{k}} \\ x & y & z \\ p _{x} & p _{y} & p _{z} \\ \end{vmatrix} $

As motion is in x-y plane (z = 0 and $ P _{z}=0 $ ), so $ \overrightarrow{L}=\overrightarrow{k}(xp _{y}-yp _{x}) $

Here x = vt, y = b, $ p _{x}=mv $ and $ p _{y}=0 $ \ $ \overrightarrow{L}=\overrightarrow{k}[ vt\times 0-bmv ]=-mvb\hat{k} $



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