A particle of mass 2 kg moves with an initial velocity $v = (4 ˇ i + 4 ˇ j) m / s$ . A constant force of $F = - 20 ˇ j N$ in applied on the particle. Initially, the particle was at (0,0). The x-coordinate of the particle when its y-coordinate again becomes zero is given by.

1.2 m

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4.8 m

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6.0 m

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3.2 m

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Solution

The correct option is C 3.2 m
Force is acting $F = - 20^j$
acceleration $a = \frac{F}{m} = \frac{- 20}{2} = - 10 m s^{- 2}^j$
Let t be the time, y cordinate become zero
by using $2^{n d}$ eq of kinetic motion
$s = u t + \frac{1}{2} a t^{2}$
s=0 $\Rightarrow 4 t + \frac{1}{2} (- 10) t^{2} = 0$
$\Rightarrow t = 0.8 s e c$
x coordinate is $4^i \times 0.8 = 3.2 m$
Heance the correct option is D

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Similar questions

Q. A particle of mass

2 kg

initially at origin

(0, 0)

moves with an initial velocity

\to V = 4^i + 4^j m/s

. If a constant force

\to F = - 20^j N

is applied on the particle, the

x -

coordinate of the particle when its

y -

coordinate again becomes zero is

A particle of mass 2 kg moves with an initial velocity of $\to v = 4^i + 4^j m s^{- 1}$ . A constant force of $\to F = - 20^j N$ is applies on the particle. Initially, the particle was at (0,0). The x-cordinate of the particle when its y-coordinate again becomes zero is given by