Motion in a Plane

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New answer posted•a year ago

Motion in a Plane Class 11th

The magnitude of vectors $\vec{O} A, \vec{O} B a n d \vec{O} C$  in the given figure are equal. The direction of $\vec{O} A + \vec{O} B - \vec{O} C$  with x-axis will be:-

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Let’s say $| \vec{O A} | = | \vec{O B} | = | \vec{O C} | = l$

$\vec{O A} = l c o s 30 ° \hat{i} + l s i n 30 ° \hat{j}$

$\vec{O A} + \vec{O B} - \vec{O C}$

= $[l c o s 30 ° \hat{i} + l s i n 30 ° \hat{j}] + [l c o s 60 ° \hat{i} + l s i n 60 ° (- \hat{j})] - [l c o s 45 ° (- \hat{i}) + l s i n 45 ° (+ \hat{j})]$

$= l (\frac{\sqrt[]{3}}{2} + \frac{1}{2} + \frac{1}{\sqrt[]{2}}) \hat{i} + l [\frac{1}{2} - \frac{\sqrt[]{3}}{2} - \frac{1}{\sqrt[]{2}}] \hat{j}$

$t a n θ = \frac{l [\frac{1}{2} - \frac{\sqrt[]{3}}{2} - \frac{\sqrt[]{2}}{2}]}{l [\frac{\sqrt[]{3}}{2} + \frac{1}{2} + \frac{\sqrt[]{2}}{2}]} = \frac{1 - \sqrt[]{3} - \sqrt[]{2}}{1 + \sqrt[]{3} + \sqrt[]{2}}$

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New answer posted•a year ago

Motion in a Plane Class 11th

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Two identical balls A and B thrown with same velocity ‘u’ at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then  $R = 4 \sqrt[]{h_{1} h_{2}}$

Reason R : Product of said heights.

$h_{1} h_{2} = (\frac{u^{2} s i n^{2} θ}{2 g}) . (\frac{u^{2} c o s^{2} θ}{2 g})$

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

First angle, 1 = $R = \frac{u^{2} s i n 2 ?}{g}$ $\frac{^{}}{}$

Another angle (2 = 90 - ) for which range will be

Same as that of 1 =

$R^{1} = \frac{u^{2} s i n 2 (90 ? ?)}{g} = \frac{u^{2}}{g} s i n 2 ? = R$

at $?_{1} = ?, ? ? ? ? ? h_{1} = \frac{u^{2} s i n^{2} ?}{2 g}$ $_{}_{} \frac{^{}^{}}{}$

$& ? ? ? ?_{2} = 90 ? ?, h_{2} = \frac{u^{2} s i n^{2} ?_{2}}{2 g} = \frac{u^{2} c o s^{2} ?}{2 g}$

$? h_{1} h_{2} = \frac{1}{4^{2}} {[\frac{u^{2} s i n 2 ?}{g}]}^{2}$

$h_{1} h_{2} = \frac{1}{4^{2}} R^{2}$

$R = 4 \sqrt[]{h_{1} h_{2}}$

So, Both statement is true & Reason is correct explanation for statement 1.

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New answer posted•a year ago

Motion in a Plane Class 11th

Which one of the following relations is true for two unit vector $\hat{A} a n d \hat{B}$  making an angle θ to each other?

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$| \hat{A} + \hat{B} | = \sqrt[]{A^{2} + B^{2} + 2 A B c o s θ}$ $^{}^{}$

$= \sqrt[]{1 + 1 + 2 c o s θ}$

$= \sqrt[]{2 (1 + c o s θ)}$

= $\sqrt[]{2 \times 2 c o s^{2} \frac{θ}{2}}$ $^{} \frac{}{}$

$| \hat{A} + \hat{B} | = 2 c o s \frac{θ}{2}$ -(1)

$| \hat{A} - \hat{B} | = \sqrt[]{A^{2} + B^{2} - 2 A B c o s θ}$

$= \sqrt[]{1 + 1 - 2 c o s θ}$

$= 2 s i n \frac{θ}{2}$ -(2)

(2) $\div$ (1)

$\frac{| \hat{A} - \hat{B} |}{| \hat{A} + \hat{B} |} = \frac{2 s i n \frac{θ}{2}}{2 c o s \frac{θ}{2}}$

$\Rightarrow | \hat{A} - \hat{B} | = | \hat{A} + \hat{B} | t a n \frac{θ}{2}$

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New answer posted•a year ago

Motion in a Plane Class 11th

$\vec{A}$ is a vector quantity such that  $| \vec{A} |$ = non-zero constant. Which of the following expression is true for $\vec{A}$ ?

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\vec{A} \times \vec{A} = A A s i n θ \hat{n}$

$= A A s i n 0 ° \hat{n}$

= 0 [since Angle between the vectors are zero degree]

$\vec{A} \times \vec{A} = 0$

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New answer posted•a year ago

Motion in a Plane Class 11th

A projectile is launched at an angle ‘α’ with the horizontal with a velocity 20 ms^-¹. After 10 s, it inclination with horizontal is β. The value of tanβ will be: (g = 10ms^-²).

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_{y} = 20 s i n α, V_{y} = v s i n β$

$V_{y} = V_{y} + a_{y} t$

$V s i n β = 20 s i n α - 10 \times 10$ ……. (ii)

$\Rightarrow (i i) \div (i)$

$\Rightarrow t a n β = t a n α - 5 s e c α$

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New answer posted•a year ago

Motion in a Plane Class 11th

A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60m, then the magnitude of displacement would be:

(Given $c o s 135 °$ = -0.7)

14 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

d = R

$60 = R [\frac{3 π}{4}]$

$R = \frac{60 \times 4}{3 π} = \frac{80}{π} m$

Displacement = $\sqrt[]{R^{2} + R^{2} - 2 R^{2} c o s 135}$ $^{}^{}^{}$

= $\sqrt[]{2 R^{2} - 2 R^{2} (- 0.7)}$ $^{}^{}$

= $\sqrt[]{3.4 R^{2}}$ $^{}$

= $\sqrt[]{3.4 {(\frac{80}{4})}^{2}}$ ${\frac{}{}}^{}$

47 m

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New answer posted•a year ago

Motion in a Plane Class 11th

The sum of two forces $\vec{P}$ and $\vec{Q}$ is $\vec{R}$ such that $| \vec{R} | = | \vec{P} |$ . The angle $θ$ (in degrees) that the resultant of $2 \vec{P}$ and $\vec{Q}$ will make with $\vec{Q}$ is,

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$| \overset{?}{P} + \overset{?}{Q} | = | \overset{?}{P} |$

^{$P^{2} + Q^{2} + 2 P Q c o s ? θ = P^{2}$}

$\Rightarrow Q + 2 P c o s ? θ = 0$

$\Rightarrow c o s ? θ = - \frac{Q}{2 P}$

$t a n ? α = \frac{2 P s i n ? θ}{2 P c o s ? θ + Q} = \infty$

$? [2 P c o s ? θ + Q = 0]$

α = 90^{?}

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New answer posted•a year ago

Motion in a Plane Class 11th

Two inclined planes are placed as shown in figure. A block in Projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t $t (\sqrt[]{2} + 1) s$ . The value of t is__________.

(use g = 10 m/s²)

17 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

From conservation of Energy

$\frac{1}{2} m v_{0}^{2} = m g h$

$v_{0} = 10 \sqrt[]{2}$

For A to B

$a = - g s i n 45 ° = \frac{- 10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l} = 2 \sqrt[]{2} + 2 \\ = 2 (\sqrt[]{2} + 1) \end{array}$

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New answer posted•a year ago

Motion in a Plane Class 11th

An object is projected in the air with initial velocity u at an angle q. The projectile motion is such that the horizontal range R, in maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be_______degree.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$R = \frac{u^{2} s i n (2 \times 45 °)}{g} = \frac{u^{2}}{g}$

$\frac{R}{2} = \frac{u^{2}}{2 g} = \frac{u^{2} s i n 20}{g}$

$s i n 2 θ = \frac{1}{2}$

$\Rightarrow 2 θ = 30 ° \Rightarrow θ = 15 °$

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New answer posted•a year ago

Motion in a Plane Class 11th

If the projection of $2 \hat{i} + 4 \hat{j} - 2 \hat{k} o n \hat{i} + 2 \hat{j} + α \hat{k}$ is zero. Then, the value of will be________.

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

${\vec{r}}_{1} = 2 \hat{i} + 4 \hat{j} - 2 \hat{k}, {\vec{r}}_{2} = \hat{i} + 2 \hat{j} + α \hat{k}$

Projection of ${\vec{r}}_{1}$ $_{}$ on ${\vec{r}}_{2}$ $_{}$ is zero $\Rightarrow {\vec{r}}_{1} . {\vec{r}}_{2}$ $_{}_{}$ = 0

2 + 8 - 2 = 0=> α= 5

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Motion in a Plane

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