At high altitude, a body explodes which is at rest into two equal fragments with one fragment receiving horizontal velocity of 10 m ⋅ s -1. Time taken by the two radius vectors connecting point of explosion to fragments to make 90° is

Question

At high altitude, a body explodes which is at rest into two equal fragments with one fragment receiving horizontal velocity of 10 m ⋅ s -1. Time taken by the two radius vectors connecting point of explosion to fragments to make 90° is (A) 10 s (B) 4 s (C) 2 s (D) 1 s

Tardigrade Admin · Accepted Answer

The given situation is shown in the following figure.

According to question, angle between two radius vector is

9 0^{\circ}

.
i.e.,

θ + θ = 9 0^{\circ}

\Rightarrow θ = 4 5^{\circ}

∴

Time taken by the two radius vectors connecting of explosion to fragments is

t

second, then
According to conservation of linear momentum,

m_{1} v_{1} = m_{2} v_{2}

\Rightarrow m v_{1} = m v_{2}

\Rightarrow v_{1} = v_{2}

Since,

v_{1}

and

v_{2}

are perpendicular to each other, hence

v_{1} \cdot v_{2} = 0

(- 10 \hat{i} - g \hat{j}) \cdot 10 \hat{i} - g \hat{j} = 0

\Rightarrow - 100 + g^{2} t^{2} = 0

\Rightarrow x^{2} = \frac{100}{g ^{2}}

= \frac{100}{1 0 ^{2}} = 1 s

Q. At high altitude, a body explodes which is at rest into two equal fragments with one fragment receiving horizontal velocity of 10m⋅s−1. Time taken by the two radius vectors connecting point of explosion to fragments to make 90∘ is

10 s

4 s

2 s

1 s

Q. At high altitude, a body explodes which is at rest into two equal fragments with one fragment receiving horizontal velocity of $10 m \cdot s^{- 1}$ . Time taken by the two radius vectors connecting point of explosion to fragments to make $9 0^{\circ}$ is