SATHEE: Kinematics Question 262

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) $m v b \hat{k}$

B) $- m v b \hat{k}$

C) $m v b \hat{i}$

D) $m v \hat{i}$

Show Answer

Answer:

Correct Answer: B

Solution:

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

As motion is in x-y plane (z = 0 and $P_{z} = 0$ ), so $\vec{L} = \vec{k} (x p_{y} - y p_{x})$

Here x = vt, y = b, $p_{x} = m v$ and $p_{y} = 0$ \ $\vec{L} = \vec{k} [v t \times 0 - b m v] = - m v b \hat{k}$

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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:

Answer:

Solution:

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