Ask question

Ask question

All categories

chubhunter [2.5K]

3 years ago

9

Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th

e mass of the eight ball, m2, is 6 g with a velocity of 1 m/s. After the balls collide, they bounce off each other. The number three ball moves off with a velocity of 5 m/s. What is the final velocity and direction of the eight ball? 8. 6 m/s 5. 7 m/s â€"5. 7 m/s â€"8. 6 m/s.

1 answer:

Ymorist [56]3 years ago

4 0

This question involves the concepts of the law of conservation of momentum and velocity.

The velocity of the eight ball is "5.7 m/s".

According to the law of conservation of momentum:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = mass of number three ball = 5 g

m₂ = mass of the eight ball = 6 g

u₁ = velocity of the number three ball = 3 m/s

u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)

v₁ = final velocity of the three number ball = - 5 m/s

v₂ = final velocity of the eight ball = ?

Therefore,

(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

$v_2=\frac{34\ g.m/s}{6\ g}\\\\$

<u>v₂ = 5.7 m/s</u>

<u></u>

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

You might be interested in

Suppose that the collector is held at a small negative voltage with respect to the grid. Will the accelerated electrons reach th

Leona [35]

Answer:

B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.

Explanation:

In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential

so according to the given situation, the option b is correct

And, the rest of the options are wrong

3 0

3 years ago

The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a

Aleksandr [31]

Answer:

Utilization, effects

Explanation:

The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.

7 0

4 years ago

Which statement describes the distribution of charge in an atom? (1) A positively charged nucleus is surrounded by one or more n

pantera1 [17]

Answer: option B

Explanation: when a neutral atom loses an electron or gains a positive charge electron, it becomes a positive ion (positively charged) and when an neutral atom gains an electronic charge or losses a positive charge electron, it becomes a negative ion (negatively charged).

4 0

3 years ago

Read 2 more answers

A monkey has a bit of a heavy for on the gas pedal. As soon as the light turns green the monkey pushes the gas pedal to the floo

Andrei [34K]

Answer:

$s=6.86m/s^2$

Explanation:

Hello,

In this case, considering that the acceleration is computed as follows:

$a=\frac{v_{final}-v_{initial}}{t}$

Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:

$a=\frac{28.82m/s-0m/s}{4.2s}\\ \\s=6.86m/s^2$

Regards.

3 0

3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago