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rosijanka [135]
3 years ago
8

A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo

ves with a speed equal to one- third the original speed of the 3.00 kg mud ball. what is the mass of the second mud ball?
Physics
1 answer:
viva [34]3 years ago
7 0
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

m_{1} * v_{1} +  m_{2} * v_{2} =  m_{1} * v'_{1} +  m_{2} * v'_{2}

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.
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Answer:

<u>r < a:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}

<u>r = a:</u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>a < r < b:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>r = b:</u>

E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

<u>r < c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

Explanation:

Gauss' Law will be applied to each region to find the E-field.

\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}

Applying Gauss' Law:

E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u />E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}<u />

<u>At r = b:</u>

<u />E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}<u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}

<u>At r = c:</u>

E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}

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