A bullet with a mass of 10 g, flying horizontally at a speed of 400 m / s, hits the middle of a ball with a mass of 1 kg

| April 4, 2021 | education

A bullet with a mass of 10 g, flying horizontally at a speed of 400 m / s, hits the middle of a ball with a mass of 1 kg, hanging on a weightless inextensible rod, and pierces it. Find the speed of the bullet after breaking the ball if the ball has risen to a maximum height of 0.2 m.

mp = 10 g = 0.01 kg.

Vp1 = 400 m / s.

msh = 1 kg.

h = 0.2 m.

g = 10 N / kg.

Vp2 -?

For a bullet with a ball, the law of conservation of total mechanical energy is valid: mp * Vp1 ^ 2/2 = mw * h * g + mp * Vp2 ^ 2/2.

mp * Vp2 ^ 2/2 = mp * Vp1 ^ 2/2 – mw * h * g.

Vp2 ^ 2 = 2 * (mp * Vp1 ^ 2/2 – mw * h * g) / mp.

The formula for determining the speed of a bullet after breaking the ball Vp2 will have the form: Vp2 = √ (2 * (mp * Vp1 ^ 2/2 – mw * h * g) / mp).

Vp2 = √ (2 * (0.01 kg * (400 m / s) ^ 2/2 – 1 kg * 0.2 m * 10 N / kg) / 0.01 kg) = 399.5 m / s.

Answer: the speed of the bullet after breaking through the ball will become Vp2 = 399.5 m / s.



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