answers for (b), (c), and (d) and explanation please! From a large distance away
ID: 1528405 • Letter: A
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answers for (b), (c), and (d) and explanation please!
From a large distance away, a particle of mass m_1 and positive charge q_1 is fired at speed v in the positive x direction straight toward a second particle, originally stationary but free to move, with mass m_2 and positive charge q_2. Both particles are constrained to move only along the x axis. (Use any variable or symbol stated above along with the following as necessary: k_e for 1/4 pi epsilon_0.) At the instant of closest approach, both particles will be moving at the same speed. Find this speed. v_c = Find the distance of closest approach. r_c = After the interaction, the particles will move far apart again. At this time, find the velocity of the following. the particle of mass m_1 vector V_m1 = the particle of mass m_2 vector V_m2 =Explanation / Answer
a.) Using conservation of momentum, the initial momentum = the final momentum (momentum at closest approach)
So, m1 v = (m1 + m2 ) vc
vc = m1v / ( m1 + m2 )
b.) Initial Kinetic Energy = 0.5 m1v2
Initial Potential energy = 0 (since they are far apart)
At the closest approach, Kinetic Energy = 0.5 ( m1 + m2 ) vc2
Potential energy at the closest approach be P
According to the conservation of energy, total initial energy = total final energy
0.5 m1v2 + 0 = 0.5 ( m1 + m2 ) vc2 + P
0.5 [ m1 (v2 - vc2 ) - m2 vc2 ] = P
if r is the distance of closest approach,
0.5 [ m1 (v2 - vc2 ) - m2 vc2 ] = ke q1q2 / r
r = 2 keq1q2 / [ m1 (v2 - vc2 ) - m2 vc2 ]
c.) & d.) Let v1 and v2 be the final velocities.
according to momentum conservation : m1v = m1v1 + m2v2 eqn 1
Also, since they move far apart again, the Potential energy will become Zero again.
Hence according to energy conservation, 0.5 m1v2 = 0.5 m1v12 + 0.5 m2v22 eqn 2
Solving eqn 1 and eqn 2 for v1 and v2
m1 ( v2 - v12 ) = m2v22 eqn 2
v2 = m1 ( v - v1 ) / m2 eqn 1
eqn 1 in eqn 2
m1 ( v2 - v12 ) / m2 = m12( v - v1 )2/ m22
( v - v1 ) ( v + v1 ) = m1 ( v - v1 )2/ m2
( v + v1 ) = m1 ( v - v1 )/ m2
v1 (1 + m1 / m2 ) = v ( m1 - 1 )
v1 = m2 (m1 - 1) v / ( m1 + m2 )
v2 = m1 ( v - v1 ) / m2 eqn 1
v2 = m1 ( v - [ m2 (m1 - 1) v / ( m1 + m2 ) ] ) / m2
v2 = (m1 / m2 ) v [ 1 - [ m2 (m1 - 1) / ( m1 + m2 ) ] ]
v2 = (m1 / m2 ) v [ m1 + m2 - m2 m1 + m2 ] / ( m1 + m2 )
v2 = m1 [ m1 + 2m2 - m2 m1 ] v / m2 ( m1 + m2 )
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