A ballistic pendulum is a device for measuring bullet speeds. One of the simplest versions consists of a block of wood hanging from two long cords. (Two cords are used so that the bottom face of the block remains parallel to the floor as the block swings upward.) A 9.4-g bullet is fired into a ballistic pendulum in which the block has an inertia of 5.5 kg , and the block rises 60 mm above its initial position.Part AWhat is the speed of the bullet just before it hits the block?Part BHow much energy is dissipated in the collision?

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aristocles aristocles · Answer 1 · 2019-11-12T01:50:12+01:00

Answer:

Part a)

[tex]v = 636 m/s[/tex]

Part b)

[tex]\Delta E = 1898 J[/tex]

Explanation:

Part a)

Let the bullet is moving initially with speed v

so by momentum conservation we will have

[tex]mv = (M + m) v_f[/tex]

[tex]v_f = \frac{0.0094 v}{0.0094 + 5.5}[/tex]

[tex]v_f = 1.706 \times 10^{-3} v[/tex]

now by energy conservation we know that

[tex]\frac{1}{2}(M + m)(v_f^2) = (M + m)gH[/tex]

[tex]H = \frac{v_f^2}{2g}[/tex]

[tex]0.06 = \frac{(1.706\times 10^{-3}v)^2}{2(9.81)}[/tex]

[tex]0.06 = 1.48 \times 10^{-7} v^2[/tex]

[tex]v = 636 m/s[/tex]

Part b)

Energy loss of the system is given as

[tex]\Delta E = \frac{1}{2}mv_i^2 - \frac{1}{2}(M + m)v_f^2[/tex]

[tex]\Delta E = \frac{1}{2}(9.4 \times 10^{-3})(636^2) - \frac{1}{2}(0.0094 + 5.5)(1.08^2)[/tex]

[tex]\Delta E = 1901.13 - 3.21[/tex]

[tex]\Delta E = 1898 J[/tex]