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2 years ago

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A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f

irst piece flies off with a speed of 6.0 m/s, and the 3rd piece flies off at 3.0 m/s what is the speed of the Second fragments? (All fragments are in the plane.)

1 answer:

RSB [31]2 years ago

3 0

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break fly off = 120°

α = β = γ = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

$\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}$

A,B and C is momentum of the fragments

$\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}$

4 x V₂ = 2 x 3

V₂ = 1.5 m/s

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A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".

Alex787 [66]

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

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Part e)

$v = 2.1 m/s$

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Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

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so the maximum height reached by the pendulum is given as

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now gravitational potential energy of the pendulum is given as

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$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

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so we have

$mgh = \frac{1}{2}mv^2$

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Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

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Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

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Part e)

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Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$

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