JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 3)
The time dependence of the position of a particle of mass m = 2 is given by $$\overrightarrow r \left( t \right) = 2t\widehat i - 3{t^2}\widehat j$$
. Its angular
momentum, with respect to the origin, at time t = 2 is
36 $$\widehat k$$
- 48 $$\widehat k$$
$$ - 34\left( {\widehat k - \widehat i} \right)$$
$$48\left( {\widehat i + \widehat j} \right)$$
Explanation
$$\overrightarrow v = 2\widehat i - 6 + \widehat j$$
At t = 2
$$\overrightarrow v = 2\widehat i - 12\widehat j$$
$$\overrightarrow P = m\overrightarrow v = 4i - 24\widehat j$$
At t = 2
$$\overrightarrow r = 4\widehat i - 12\widehat j$$
$$\overrightarrow L = \overrightarrow r \times \overrightarrow P = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 4 & { - 12} & 0 \cr 4 & { - 24} & 0 \cr } } \right|$$
$$ = \left\{ {4( - 2) + 4 \times 12} \right\}\widehat k$$
$$ = \left( { - 96 + 48} \right)\widehat k$$
$$ = \left( - \right)48\widehat k$$
At t = 2
$$\overrightarrow v = 2\widehat i - 12\widehat j$$
$$\overrightarrow P = m\overrightarrow v = 4i - 24\widehat j$$
At t = 2
$$\overrightarrow r = 4\widehat i - 12\widehat j$$
$$\overrightarrow L = \overrightarrow r \times \overrightarrow P = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 4 & { - 12} & 0 \cr 4 & { - 24} & 0 \cr } } \right|$$
$$ = \left\{ {4( - 2) + 4 \times 12} \right\}\widehat k$$
$$ = \left( { - 96 + 48} \right)\widehat k$$
$$ = \left( - \right)48\widehat k$$
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