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3 years ago

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A very big hockey puck with mass 2.5 kg traveling 20 degrees north of east at 10.0 m/s strikes a puck with a mass of 4.0 kg trav

eling north at 2 m/s. The 2.5 kg puck exits the collision in a direction that is 30 deg. north of east at a velocity of 8.0 m/s. What is the magnitude and direction of the 4.0 kg puck's velocity?

1 answer:

snow_tiger [21]3 years ago

4 0

Answer:

Velocity = 0.47 m/s

Direction = 65 degrees South of east.

(295 degrees counter clockwise from + X axis).

Explanation:

Conservation of momentum along the X and Y directions can be used to determine the velocity of the 4 kg puck.

Along the X direction momentum conservation is as follows:

2.5 cos 20 + 4.0 cos 90 = 2.5 cos 30 + 4 v cos α

⇒ v cos α = 2.5 ( cos 20 - cos 30) ÷ 4 = 0.46 m/s

2.5 sin 20 + 4 sin 90 = 2.5 sin 30 + 4 v sin α

v sin α = (2.5 sin 20 - 2.5 sin 30) ÷ 4 = -0.0987 m/s

v = √(v² cos² α + v² sin² α) = √ (0.46² + (-0.0987)² = 0.47 m/s

Direction = α = tan⁻¹ (-0.0987/0.46) = -65 degrees = 65 degrees clock wise from +X axis or 295 degrees counter clockwise from +X axis

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d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

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Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

$\texttt{ }$

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

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<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

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$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<h3>Step B:</h3>

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

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<em>Next we will calculate the work done on the gas:</em>

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$W_B = -2.00(5 - 3\frac{1}{3})$

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<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

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$\boxed{W \approx -743 \textt{ J}}$

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$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

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<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
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$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

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