CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Acceleration in 2D

Two particles...

Question

Two particles having position vectors $\to r = (3 \to i + 5 j) m$ and ${\to r}_{2} = (- 5 i + 3 j) m$ are moving with velocities ${\to V}_{1} = (4^i - 4^j) m s^{- 1}$ and ${\to V}_{2} = (a^i - 3^j) m s^{- 1}$ . If they collide after $2$ seconds , the value of $a$ is

A

-

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

-

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

-

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

-

8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D - $8$
Let us consider the values as following:
$\begin{matrix} r_{1} = (3 i + 5 j) m r_{2} = (- 5 i + 3 j) m v_{1} = (4 i - 4 j) m / s a n d l e t v_{2} b e = (- a i - 3 j) m / s t = 2 sec \end{matrix}$
If two particles collides after time $" t "$ they travel equal distance.
$\begin{matrix} r_{1} + v_{1} t = r_{2} + v_{2} t (3 i + 5 j) + (4 i - 4 j) \times 2 = - 5 i + 3 j + (- a i - 3 j) 2 o n s o l v i n g w e g e t, 3 i + 8 i + 5 j - 8 j = - 5 i + 3 j - 2 a i - 6 i 11 i + 5 i - 3 j = - 2 a i - 3 j 16 i = - 2 a i a = \frac{- 16}{2} = - 8 m / s^{2} \end{matrix}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Two particles

1

2

are moving with constant velocities

{\to v}_{1} = 4^i - 3^j m/s

{\to v}_{2} = b^i -^j m/s

respectively. The position vectors of the particles at time

t = 0

{\to r}_{1} = 5^i + 2^j m

{\to r}_{2} = - 4^i - 4^j m

. If they collide, which of the following is correct. (

b

Q. Two particles having position vectors

_{1} = 3^i + 5^j

_{2} = - 5^i + 3^j

m are moving with velocities

_{1} = 4^i + 3^j

_{2} = - a^i + 4^j

m/s. If they collide after

2

seconds, the value of

a

Q. Two particles having position vectors

_{1} = (3^i + 5^j) m

_{2} = (- 5^i + 3^j) m

are moving with velocities

_{1} = (4^i - 4^j) {ms}^{- 1}

_{2} = (a^i - 3^j) {ms}^{- 1}

. If they collide after

2 s

, the value of the constant '

a

Q. Two particles having the position

{¯ r}_{1} = (3^i + 5^j) m

{¯ r}_{2} = (- 5^i - 3^j) m

move with velocities

{¯ V}_{1} = (4^i + 3^j) m / s

{¯ V}_{2} = (a^i + 7^j) m / s

. If the particles collide, then value of a must be :

Two particles of masses $m_{1} a n d m_{2}$ in projectile motion have velocities ${\to v}_{1} a n d {\to v}_{2}$ respectively at time t = 0 and that is the moment when they collide. Their velocities become ${\to v}_{1}^{'} a n d {\to v}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of magnitude of change in total momentum would be

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Acceleration

PHYSICS

Watch in App

explore_more_icon

Explore more

Acceleration in 2D

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon