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lutik1710 [3]
3 years ago
11

Two identical billiard balls are rolling toward each other at the same speed. What will be true after they collide head–on? They

will both stop rolling immediately after the collision. One ball will stop rolling as the other bounces backward at a slower speed. They will both bounce back at a faster speed after the collision. They will both bounce back at the same speed they had before the collision.
Physics
2 answers:
Papessa [141]3 years ago
8 0

Answer:

Explanation:

One ball will stop rolling as the other bounces backward at a slower speed. They will both bounce back at a faster speed after the collision.

mylen [45]3 years ago
5 0
<span> They will both stop rolling immediately after the collision.</span>
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2 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

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