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3 years ago

8

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

fired with a speed of 619 m/s at a block of wood and passes completely through it. The speed of the block is 22 m/s immediately after the bullet exits the block.
Required:
Determine the speed of the bullet as it exits the block.

1 answer:

kozerog [31]3 years ago

4 0

Answer:

v’= 279.66 m / s

Explanation:

We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.

Initial instant. Before the crash

p₀ = m v₀ + 0

Final moment. After the crash

p_f = m v + M v ’

how the tidal wave is preserved

p₀ = p_f

m v₀ = m v + M v ’

v = $\frac{m v_o - Mv'}{m}$

let's calculate

v ’= $\frac{0.00467 \ 619 - 0.072 \ 22}{0.004676}$

v ’= $\frac{2.89- 1.584}{ 0.00467}$

v ’= 279.66 m / s

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Explanation:

Part a

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The stopping sight distance is given as

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Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
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Substituting values

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So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

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Answer:

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