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3 years ago

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A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s. it strikes a second hockey puck with a m

ass 0.11 kg. the first hockey puck comes to rest after the collision. What is the velocity of the second hockey puck after the collision? ( Round your answer to the nearest integer.)

1 answer:

Igoryamba3 years ago

3 0

Answer:

$v_2 = 23.182m/s$

Explanation:

Given

$m_1 = 0.17kg$

$m_2 = 0.11kg$

$u_1 = 15m/s$ -- Initial Velocity of the first

$u_2 = 0m/s$ -- Initial Velocity of the second

$v_1 = 0m/s$ -- Final Velocity of the first

Required

Determine the final velocity of the second hockey puck ( $v_2$ )

This question illustrates law of conservation of momentum and it'll be solved using the following formula:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ ---- law of conservation of momentum

Substitute in the right values

$0.17 * 15 + 0.11*0 = 0.17 * 0 + 0.11*v_2$

$0.17 * 15 + 0 = 0 + 0.11*v_2$

$2.55 = 0.11*v_2$

Solve for v2

$v_2 = 2.55/0.11$

$v_2 = 23.182m/s$

<em>Hence, the final velocity of the second hockey puck is 23.182m/s</em>

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Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

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That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

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$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

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$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

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$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

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$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

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Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.

We need to find the distance between the plates.

At equilibrium,

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Since, E = V/r (r is distance between plates)

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