A proton (mass m = 1.67 10-27 kg) is being accelerated along a straight line at 3.1 1015 m/s2 in a machine. The proton has an initial speed of 2.4 107 m/s and travels 3.6 cm.
(a) What is its speed?
m/s
(b) What is the increase in its kinetic energy?
J
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To solve this problem, we can use the equations of motion and the principle of conservation of energy.
(a) To find the final speed of the proton, we can use the equation of motion:
v_final^2 = v_initial^2 + 2a*s
Where:
v_final is the final speed,
v_initial is the initial speed,
a is the acceleration,
s is the displacement.
Substituting the given values:
v_final^2 = (2.4 * 10^7 m/s)^2 + 2(3.1 * 10^15 m/s^2)(3.6 * 10^-2 m)
Calculating:
v_final^2 = 5.76 * 10^14 m^2/s^2 + 22.464 m^2/s^2
Adding the two values:
v_final^2 = 5.76 * 10^14 m^2/s^2 + 22.464 m^2/s^2 = 5.76 * 10^14 m^2/s^2
Now, taking the square root of both sides:
v_final = sqrt(5.76 * 10^14 m^2/s^2)
v_final = 2.4 * 10^7 m/s
The final speed of the proton is 2.4 * 10^7 m/s.
(b) To find the increase in kinetic energy, we need to use the principle of conservation of energy. The initial kinetic energy (K1) is given by the equation:
K1 = (1/2)mv_initial^2
Where:
m is the mass of the proton,
v_initial is the initial speed.
Substituting the given values:
K1 = (1/2)(1.67 * 10^-27 kg)(2.4 * 10^7 m/s)^2
Calculating:
K1 = (1/2)(1.67 * 10^-27 kg)(5.76 * 10^14 m^2/s^2)
K1 = 0.4 * (1.67 * 5.76) * 10^-13 kg * m^2/s^2
K1 = 3.349 * 10^-13 kg * m^2/s^2
The initial kinetic energy of the proton is 3.349 * 10^-13 J.
The final kinetic energy (K2) is given by the equation:
K2 = (1/2)mv_final^2
Substituting the given values:
K2 = (1/2)(1.67 * 10^-27 kg)(2.4 * 10^7 m/s)^2
Calculating:
K2 = (1/2)(1.67 * 10^-27 kg)(5.76 * 10^14 m^2/s^2)
K2 = 0.4 * (1.67 * 5.76) * 10^-13 kg * m^2/s^2
K2 = 3.349 * 10^-13 kg * m^2/s^2
The final kinetic energy of the proton is 3.349 * 10^-13 J.
The increase in kinetic energy is given by the difference between the final and initial kinetic energies:
Increase in kinetic energy = K2 - K1
Increase in kinetic energy = (3.349 * 10^-13 J) - (3.349 * 10^-13 J)
Increase in kinetic energy = 0 J
Therefore, the increase in kinetic energy of the proton is 0 J.