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4 years ago

7

Two blocks slide without friction. Block 1 has a mass of 1.6 kg and a velocity of +5.5 m/s. Block 2 has a mass of 2.4 kg and a v

elocity of +2.5 m/s. After the collision, block 2 has a velocity of +4.9 m/s. What is the velocity of block 1 after the collision?

1 answer:

fomenos4 years ago

6 0

Answer:

The velocity of block 1 after the collision is 1.9 m/s.

Explanation:

Hi there!

The momentum of the system composed by the two blocks is conserved (i.e. it remains constant) because no external force is acting on the blocks at the moment of the collision. Then, the momentum of the system before the collision is equal to the momentum after the collision. The momentum of the system is calculated adding the momenta of the two blocks.

initial momentum of the system = final momentum of the system

p1 + p2 = p1´ + p2´

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

p1 = initial momentum of block 1.

p2 = initial momentum of block 2.

p1´ = final momentum of block 1.

p2´ = final momentum of block 2.

m1 = mass of block 1.

m2 = mass of block 2.

v1 = initial velocity of block 1 (before the collision).

v2 = initial velocity of block 2.

v1´ = final velocity of block 1.

v2´ = final velocity of block 2.

Let´s write the equation with the data we have:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

1.6 kg · 5.5 m/s + 2.4 kg · 2.5 m/s = 1.6 kg · v1´ + 2.4 kg · 4.9 m/s

Solving for v1´:

14.8 kg · m/s = 1.6 kg · v1´ + 11.76 kg · m/s

(14.8 kg · m/s - 11.76 kg · m/s) / 1.6 kg = v1´

v1´ = 1.9 m/s

The velocity of block 1 after the collision is 1.9 m/s.

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The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

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$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

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(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

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(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

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(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

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