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IrinaVladis [17]
4 years ago
9

To understand how to find the velocities of objects after a collision.

Physics
1 answer:
trasher [3.6K]4 years ago
5 0

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = \frac{v}{2}

Part C: v_1 = \frac{v}{3}; v_2 = \frac{4v}{3}

Part D: v_1 = v_2 = \frac{v}{4}

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = \frac{1}{2}m.v^{2}

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f}  )

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i}  + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}

v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}

For all the collisions, object 2 is static, i.e. v_{2i} = 0

<u>Part A</u>: Both objects have the same mass (m), v_{1i} = v and collision is elastic:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = 0

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.m}{m+m}.v

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}

v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m.v}{m+m}

v_1 = v_2 = \frac{v}{2}

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = \frac{2m - m}{2m + m } . v

v_1 = \frac{v}{3}

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.2m}{2m+m}.v

v_2 = \frac{4v}{3}

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m}{m+3m}.v

v_1 = v_2 = \frac{v}{4}

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Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

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Explanation:

Electromagnetic radiation refers to

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X rays are the rays produced when a negatively charged electrode is heated by electricity and electrons are released, thereby producing energy. It is a type of radiation called EM waves.

Radio wave are wave from the portion of the electromagnetic spectrum at lower frequencies than microwaves. It is an EM wave.

Ultraviolet wave Invisible rays that are part of the energy that comes from the sun. It is an example of EM wave.

Cosmic ray is a high-speed particle either an atomic nucleus or an electron that travels through space.

Cosmic ray is not an electromagnetic wave.

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Cosmic ray is not considered electromagnetic radiation

learn more about electromagnetic radiation here:

<u>brainly.com/question/13695751</u>

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