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RUDIKE [14]
2 years ago
11

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. A croquet player strikes the green ball and it gains

an initial velocity of 2.26 m/s. It then strikes the blue ball, which is initially at rest. After the collision, the green ball has a velocity of 1.14 m/s in the same direction. If the balls roll on a frictionless surface and the collision is head-on, what is the final velocity of the blue ball? (Round your answer to the nearest hundredths place.)
Physics
1 answer:
Gelneren [198K]2 years ago
4 0

Answer:

v' = 1.21 m/s

Explanation:

Mass of a green ball, m = 0.525 kg

Mass of a blue ball, m' = 0.482 kg

Initial velocity of green ball, u = 2.26 m/s

Initial velocity of blue ball, u' = 0 (at rest)

After the collision,

The final velocity of the green ball, v = 1.14 m/s

We need to find the final velocity of the blue ball after the collision if the collision is head on. Let v' is the final velcity of the blue ball. Using the conservation of momentum to find it :

mu+m'u'=mv+m'v'\\\\0.525 (2.26)+0=0.525 (1.14)+0.482v'\\\\0.588=0.482v'\\\\v'=\dfrac{0.588}{0.482}\\\\v'=1.21\ m/s

So, the final velocity of the blue ball is 1.21 m/s.

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Answer:60mph

Explanation:

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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

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 this case is 0 as the road is level

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)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

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

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

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