A ball falls freely from a height of 45m

Asked by admin @ 20/08/2021 in Physics viewed by 419 People

A ball falls freely from a height of 45 m. When the ball is at a<br />height of 25 mit explodes into two equal pieces. One of them<br />moves horizontally with a speed of 10 ms. The distance<br />between the two pieces on the ground is<br />(a) 20 m (b) 30 m (c) 40 m (d) 60 m

Answered by admin @ 20/08/2021

Answer:

Option.(a) 20 m

Explanation:

A ball of mass “m” is falling freely from a height of 45 m with an initial velocity u = 0.

At 25 m the ball divides into two parts and the mass of each piece becomes “m/2” and final velocity be “Vy” in the vertical direction.

The horizontal velocity of one of the piece is 10 m/s.

We can see from the figure that the horizontal velocity for both the pieces will be the same as “Vx” but in opposite direction. Since no horizontal force is acting on to balls, therefore, momentum will be conserved.

We know,

Momentum = Mass * Velocity

We can write,

(Positive direction momentum of one piece) = (Negative direction momentum of another piece)

⇒ (m/2) * 10 = (m/2) * (-Vx)

⇒ Vx = - 10 m/s

∴ Vx relative = 10 – (-10) = 10 + 10 = 20 m/s

Also,

Final velocity, Vy = √(u + 2 gh) = √[0 + 2 * 10 * (45-25)] = 20 m/s

Now, we need to find the time that the ball will take to touch the ground from a height of 25 m. We have,

s = ut + ½ gt²

⇒ 25 = 20t + ½ * 10 t²

⇒ 5t² + 20t – 25 = 0

⇒ 5t² + 25t – 5 t – 25 = 0

⇒ 5t(t+5) – 5(t+5) = 0

⇒ (5t - 5)(t+5) = 0

⇒ t = 1 s [neglecting the negative value]

Thus,

The relative horizontal distance between the two pieces on the ground will be

= (Vx relative) * time

= 20 m/s * 1 s

= 20 m

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