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A 5.1 g bullet is fired into a 2.3 kg ballistic pendulum. The bullet emerges from the block with a speed of 221 m/s, and the block rises to a maximum height of 7 cm . Find the initial speed of the bullet. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m/s.

Respuesta :

Answer:

748.62 m /s.

Explanation:

mass of bullet m = .0051  kg .

mass of block M = 2.3 kg

block rises to a height of .07 m so velocity of block after collision

V = √ 2 gh

=√ (2 x 9.8 x .07 )

= 1.17 m /s

velocity of bullet after collision v = 221 m /s

Now we shall apply law of conservation of momentum to find out the velocity of bullet before collision .

Let it be Vx . then

5.1 x 10⁻³ x Vx + 0 = 5.1 x 10⁻³ x 221 + 2.3 x 1.17

= 1.127 + 2.691 = 3.818

Vx = 748.62 m /s

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