Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.
<h3>What is the relation between the masses of A and B?</h3>
Mass of piece B = Mb
- Velocities of pieces A and B are Va and Vb respectively.
- As per conservation of momentum,
Ma×Va = Mb×Vb
So, 1.9Mb × Va = Mb×Vb
=> 1.9Va = Vb
<h3>What are the kinetic energy of piece A and B?</h3>
- Expression of kinetic energy of piece A = 1/2 × Ma × Va²
- Kinetic energy of piece B = 1/2 × Mb × Vb²
- Total kinetic energy= 7900J
=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900
=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900
=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j
=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule
- Kinetic energy of piece B = 7900 - 2724 = 5176 Joule
Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.
Learn more about the kinetic energy here:
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<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
