JEE Main Motion in Plane PYQ

6.(JEE Main 2021 (Online) 22th July Evening Shift )

Three particles P, Q and R are moving along the vectors $\vec{A} = \hat{i} + \hat{j}$ , $\vec{B} = \hat{j} + \hat{k}$ and $\vec{C} = - \hat{i} + \hat{j}$ respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector $\vec{A}$ and $\vec{B}$ . Similarly particle Q is moving normal to the plane which contains vector $\vec{A}$ and $\vec{C}$ . The angle between the direction of motion of P and Q is $\cos^{- 1} (\frac{1}{\sqrt{x}})$ . Then the value of x is _______________.

Correct answer is (3)

${\hat{n}}_{1} \frac{\vec{A} \times \vec{B}}{| \vec{A} \times \vec{B} |} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$

${\hat{n}}_{2} \frac{\vec{A} \times \vec{C}}{| \vec{A} \times \vec{C} |} = \hat{k}$

$\cos θ = {\hat{n}}_{1} . {\hat{n}}_{2} = \frac{1}{\sqrt{3}}$

7. (JEE Main 2020 (Online) 4th September Morning Slot )

Starting from the origin at time t = 0, with initial velocity 5 $\hat{j}$ ms^-1 , a particle moves in the x-y plane with a constant acceleration of $(10 \hat{i} + 4 \hat{j})$ ms^-2. At time t, its coordinates are (20 m, y₀ m). The values of t and y₀ are, respectively:

A. 5s and 25 m

B. 2s and 18 m

C. 2s and 24 m

D. 4s and 52 m

rrect Option is (B)

$y = u_{y} t + \frac{1}{2} a_{y} t^{2}$

$y = 5 t + \frac{1}{2} (4) t^{2}$

$y = 5 t + 2 t^{2}$

and $x = 0 (t) + \frac{1}{2} (10) (t^{2}) = 20$

$t = 2 s$

$\Rightarrow y = 10 + 8 = 18 m$

8. (JEE Main 2020 (Online) 9th January Evening Slot )

A particle starts from the origin at t = 0 with an
initial velocity of 3.0 $\hat{i}$ m/s and moves in the
x-y plane with a constant acceleration $(6 \hat{i} + 4 \hat{j})$ m/s² . The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is :-

A. 40

B. 32

C. 50

D. <60 /p>

rrect Option is (D)

$\vec{u}$ = 3.0 $\hat{i}$

$\vec{a}$ = $(6 \hat{i} + 4 \hat{j})$

$\vec{S} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$

x = 3t + $\frac{1}{2} 6 t^{2}$

= 3t + 3t² .....(1)

y = $\frac{1}{2} \times 4 \times t^{2}$ = 32

$\Rightarrow$ t = 4 s .... (2)

x = 3 × 4 + 3 × 4² = 12 + 48 = 60 m

9. (JEE Main 2019 (Online) 9th April Evening Slot )

The position vector of a particle changes with time according to the relation $\vec{r} (t) = 15 t^{2} \hat{i} + (4 - 20 t^{2}) \hat{j}$
What is the magnitude of the acceleration at t = 1 ?

A. 50

B. 25

C. 40

D. 100

rrect Option is (A)

$\vec{r} = (15 t^{2}) \hat{i} + (4 - 20 t^{2}) \hat{j}$

$\vec{v} = \frac{d \vec{r}}{d t} = (30 t) \hat{i} - (40 t) \hat{j}$

$\vec{a} = \frac{d \vec{v}}{d t} = (30) \hat{i} - (40) \hat{j}$

$| \vec{a} | = 50$

10. (JEE Main 2019 (Online) 11th January Evening Slot )

A particle moves from the point $(2.0 \hat{i} + 4.0 \hat{j})$ m, at t = 0, with an initial velocity $(5.0 \hat{i} + 4.0 \hat{j})$ ms^{$-$ 1}. It is acted upon by a constant force which produces a constant acceleration $(4.0 \hat{i} + 4.0 \hat{j})$ ms^{$-$ 2}. What is the distance of the particle from the origin at time 2 s?

A. 15 m

B. $20 \sqrt{2}$ m

C. $10 \sqrt{2}$ m

D. 5 m

rrect Option is (B)

$\vec{S} = (5 \hat{i} + 4) 2 + \frac{1}{2} (4 \hat{i} + 4 \hat{j}) 4$

$= 10 \hat{i} + 8 \hat{j} + 8 \hat{i} + 8 \hat{j}$

$\vec{r_{f}} - \vec{r_{i}} = 18 \hat{i} + 16 \hat{j}$

$\vec{r_{f}} = 20 \hat{i} + 20 \hat{j}$

$| \vec{r_{f}} | = 20 \sqrt{2}$