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boyakko [2]
3 years ago
14

A rifle of mass 2 kg is suspended by strings. The rifle fires a bullet of mass 0.01 kg at a speed of 200 m/s. The recoil velocit

y of the rifle is about ____.
(A) 0.001 m/s.
(B) 0.01 m/s.
(C) 0.1 m/s.
(D) 1 m/s
Physics
1 answer:
ozzi3 years ago
5 0

Answer:

option D  

Explanation:

given,

mass of the rifle,M = 2 Kg

mass of the bullet , m = 0.01 Kg

speed of the bullet, v = 200 m/s

recoil velocity ,v'= ?

initial velocity of bullet and rifle is zero.

using conservation of energy

Mu + m u' = M v'+ m v

0 + 0  = 2 x v' + 0.01 x 200

2 v' = -2

   v' = -1 m/s

negative sign represent the velocity of rifle is in opposite direction of bullet.

hence, the correct answer is option D  

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at t=46/22, x=24 699/1210 ≈ 24.56m

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The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

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Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

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Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

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Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

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