Test Index

Impulse & Momentum

Section: Physics

Question : 7 of 16

Marks: +1, -0

M=0.2\ \mathrm{kg}

x y

(x=-l,\ y=-h)

l=10\ \mathrm{m}

h=1\ \mathrm{m}

t=0

a=10\ \mathrm{m}/\mathrm{s}^2

x

\overline{L}

\vec{\tau}

\hat{i},\ \hat{j}

\hat{k}

x,\ y

\hat{k}=\hat{i}\times\hat{j}

Solution:

Particle is initially at rest in the

x y

-plane at a point

x=-\ell,\ y=-h

where

\ell=10\ \mathrm{m}

and

h=1\ \mathrm{m}

From

s=u t+\frac{1}{2} a t^{2}

20=\frac{1}{2}\times10\times t^{2}

\therefore t=2\ \mathrm{sec}

Torque,

\vec{\tau}=\vec{r}\times\vec{F}

and here

\vec{r}_{B}=10\hat{i}-\hat{j}

\vec{F}=m\vec{a}=0.2\times10\hat{i}=2\hat{i}

\therefore \vec{\tau}=(10\hat{i}-\hat{j})\times(2\hat{i})=2\hat{k}

Angular momentum,

\vec{L}=\vec{r}_{B}\times\vec{P}=\vec{r}_{B}\times m\vec{v}

From

\vec{v}=u+\vec{a}=10\hat{i}\times2=20\hat{i}

\vec{L}=(0.2)[(10\hat{i}-\hat{j})\times20\hat{i}]=4\hat{k}

At point

A(0,-1)

Torque

\vec{\tau}=\vec{r}_{A}\times\vec{F}=(-\hat{j})\times2\hat{i}=2\hat{k}\quad[\because \vec{r}_{A}=-\hat{j}]

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