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JEE Main 2010
Exam
JEE Main
Year
2010
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Main 2010 · Physics · Motion in Two Dimensions

A small particle of mass m is projected at an angle θ with the x...

A small particle of mass mm is projected at an angle θ\theta with the x\mathrm{x}-axis with an initial velocity v0\mathrm{v}_0 in the x−y\mathrm{x}-\mathrm{y} plane as shown in the figure. At a time t<v0sin⁡θgt < \frac{v_0 \sin \theta}{g}, the angular momentum of the particle is where i^,j^\hat{\mathrm{i}}, \hat{\mathrm{j}} and k^\hat{\mathrm{k}} are unit vectors along x,y\mathrm{x}, \mathrm{y} and z\mathrm{z}-axis respectively.A small particle of mass m is projected at an angle θ with the x -axis with an initial velocity v_0 in the x-y plane as

Options

  1. A−mgv⁡0t2cos⁡θj^-\operatorname{mgv}_0 \mathrm{t}^2 \cos \theta \hat{\mathrm{j}}
  2. Bmgv⁡0tcos⁡θk^\operatorname{mgv}_0 t \cos \theta \hat{\mathrm{k}}
  3. C−12mgv0t2cos⁡θk^-\frac{1}{2} m g v_0 t^2 \cos \theta \hat{k}
  4. D12mgv0t2cos⁡θi^\frac{1}{2} m g v_0 t^2 \cos \theta \hat{i}

Correct answer

C. −12mgv0t2cos⁡θk^-\frac{1}{2} m g v_0 t^2 \cos \theta \hat{k}

Step-by-step solution

L→=m(r⃗×v→)L→=m[v0cos⁡θti^+(v0sin⁡θt−12gt2)j^]×[v0cos⁡θi^+(v0sin⁡θ−g g)j^]=mv0cos⁡θt[−12g g]k^=−12mgv0t2cos⁡θk^\begin{aligned} \overrightarrow{\mathrm{L}} & =m(\vec{r} \times \overrightarrow{\mathrm{v}}) \\ \overrightarrow{\mathrm{L}} & =m\left[\mathrm{v}_0 \cos \theta t \hat{\mathrm{i}}+\left(\mathrm{v}_0 \sin \theta \mathrm{t}-\frac{1}{2} g \mathrm{t}^2\right) \hat{\mathrm{j}}\right] \times\left[\mathrm{v}_0 \cos \theta \hat{\mathrm{i}}+\left(\mathrm{v}_0 \sin \theta-g \mathrm{~g}\right) \hat{\mathrm{j}}\right] \\ & =m \mathrm{v}_0 \cos \theta \mathrm{t}\left[-\frac{1}{2} g \mathrm{~g}\right] \hat{\mathrm{k}} \\ & =-\frac{1}{2} m g \mathrm{v}_0 \mathrm{t}^2 \cos \theta \hat{\mathrm{k}}\end{aligned}
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