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2 years ago

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An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

tially. if the mass of the second cart is m2=0.43kg, how much kinetic energy is lost as a result of the collision?

1 answer:

arsen [322]2 years ago

7 0

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

V = 0.2977 m/s

The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

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1) At x = 6.6 cm, $E_x=3.47\cdot 10^6 N/C$

2) At x = 6.6 cm, $E_y=0$

3) At x = 1.45 cm, $E_x=-3.76\cdot 10^6N/C$

4) At x = 1.45 cm, $E_y=0$

5) Surface charge density at b = 4 cm: $+62.75 \mu C/m^2$

6) At x = 3.34 cm, the x-component of the electric field is zero

7) Surface charge density at a = 2.9 cm: $+65.25 \mu C/m^2$

8) None of these regions

Explanation:

1)

The electric field of an infinite sheet of charge is perpendicular to the sheet:

$E=\frac{\sigma}{2\epsilon_0}$

where

$\sigma$ is the surface charge density

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

The field produced by a thick slab, outside the slab itself, is the same as an infinite sheet.

So, the electric field at x = 6.6 cm (which is on the right of both the sheet and the slab) is the superposition of the fields produced by the sheet and by the slab:

$E=E_1+E_2=\frac{\sigma_1}{2\epsilon_0}+\frac{\sigma_2}{2\epsilon_0}$

where

$\sigma_1=-2.5\mu C/m^2 = -2.5\cdot 10^{-6}C/m^2\\\sigma_2=64 \muC/m^2 = 64\cdot 10^{-6}C/m^2$

The field of the sheet is to the left (negative charge, inward field), while the field of the slab is the right (positive charge, outward field).

So,

$E=\frac{1}{2\epsilon_0}(\sigma_1+\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}+64\cdot 10^{-6})=3.47\cdot 10^6 N/C$

And the negative sign indicates that the direction is to the right.

2)

We note that the field produced both by the sheet and by the slab is perpendicular to the sheet and the slab: so it is directed along the x-direction (no component along the y-direction).

So the total field along the y-direction is zero.

This is a consequence of the fact that both the sheet and the slab are infinite along the y-axis. This means that if we take a random point along the x-axis, the y-component of the field generated by an element of surface dS of the sheet (or the slab), $dE_y$ , is equal and opposite to the y-component of the field generated by an element of surface dS of the sheet located at exactly on the opposite side with respect to the x-axis, $-dE_y$ . Therefore, the net field along the y-direction is always zero.

3)

Here it is similar to part 1), but this time the point is located at

x = 1.45 cm

so between the sheet and the slab. This means that both the fields of the sheet and of the slab are to the left, because the slab is negatively charged (so the field is outward). Therefore, the total field is

$E=E_1-E_2$

Substituting the same expressions of part 1), we find

$E=\frac{1}{2\epsilon_0}(\sigma_1-\sigma_2)=\frac{1}{2(8.85\cdot 10^{-12})}(-2.5\cdot 10^{-6}-64\cdot 10^{-6})=-3.76\cdot 10^6N/C$

where the negative sign indicates that the direction is to the left.

4)

This part is similar to part 2). Since the field is always perpendicular to the slab and the sheet, it has no component along the y-axis, therefore the y-component of the electric field is zero.

5)

Here we note that the slab is conductive: this means that the charges in the slab are free to move.

We note that the net charge on the slab is positive: this means that there is an excess of positive charge overall. Also, since the sheet (on the left of the slab) is negatively charged, the positive charges migrate to the left end of the slab (at a = 2.9 cm) while the negative charges migrate to the right end (at b = 4 cm).

The net charge per unit area of the slab is

$\sigma=+64\mu C/m^2$

And this the average of the surface charge density on both sides of the slab, a and b:

$\sigma=\frac{\sigma_a+\sigma_b}{2}$ (1)

Also, the infinite sheet located at x = 0, which has a negative charge $\sigma_1=-2.5\mu C/m^2$ , induces an opposite net charge on the left surface of the slab, so

$\sigma_a-\sigma_b = +2.5 \mu C/m^2$ (2)

Now we have two equations (1) and (2), so we can solve to find the surface charge densities on a and b, and we find:

$\sigma_a = +65.25 \mu C/m^2\\\sigma_b = +62.75 \mu C/m^2$

6)

Here we want to calculate the value of the x-component of the electric field at

x = 3.34 cm

We notice that this point is located inside the slab, because its edges are at

a = 2.9 cm

b = 4.0 cm

But slab is conducting , and the electric field inside a conductor is always zero (because the charges are in equilibrium): therefore, this means that the x-component of the electric field inside the slab is zero

7)

We calculated the value of the charge per unit area on the surface of the slab at x = a = 2.9 cm in part 5), and it is $\sigma_a = +65.25 \mu C/m^2$

8)

As we said in part 6), the electric field inside a conductor is always zero. Since the slab in this problem is conducting, this means that the electric field inside the slab is zero: therefore, the regions where the field is zero is

2.9 cm < x < 4 cm

So the correct answer is

"none of these region"

Learn more about electric fields:

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