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Anna35 [415]
3 years ago
9

An air track car with a mass of 0.55 kg and velocity of 5.8 m/s to the right collides and couples with a 0.45 kg car moving to t

he left with a velocity of 3.9 m/s. What is the combined velocity of the cars just after the collision?
Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

Explanation:

Given:

m1 = 0.55 kg

m2 = 0.45 kg

vi1 = +5.8 m/a

vi2 = -3.9 m/a

Solve for V

m1vi1 + m2vi2 = (m1 + m2)V

(0.55 kg)(+5.8 m/s) + (0.45 kg)(-3.9 m/s) = (0.55 + 0.45)V

V = +1.46 m/a

This means that the combined mass after the collision is moving to the right at 1.46 m/s

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A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

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r=0.070 cm =7\cdot 10^{-4} m is the distance between the charges

q_1=q_2=q since the charges are identical

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F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C

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