Question

One particle has a mass of 2.11 x 10^{-3} kg and a
charge of +8.91 μC. A second particle has a mass of 7.27 x
10^{-3} kg and the same charge. The two particles are
initially held in place and then released. The particles fly apart,
and when the separation between them is 0.128 m, the speed of the
2.11 x 10^{-3} kg-particle is 142 m/s.

Find the initial separation between the particles.

Answer #1

Solution :

Given :

m_{1} = 2.11 x 10^{-3} kg

q_{1} = + 8.91 μC

m_{2} = 7.27 x 10^{-3} kg

q_{2} = + 8.91 μC

.

Final separation (rf) = 0.128 m

v_{1} = 142 m/s

.

Since both the particles are initially at rest.

The initial momentum of these particles will be zero.

So, According to the conservation of momentum : P_{f} =
P_{i}

∴ m_{1} v_{1} + m_{2} v_{2} =
0

∴ (2.11 x 10^{-3} kg)(142 m/s) + (7.27 x 10^{-3}
kg) (v_{2}) = 0

∴ v_{2} = - 41.213 m/s

.

Now according to the conservation of energy : PE_{i} =
PE_{f} + KE_{total}

∴ PE_{i} = PE_{f} + KE_{1} +
KE_{2}

Therefore, Initial separation between the charges will be
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