A proton (mass m = 1.67 10-27 kg) is being accelerated along a straight line at 3.6 1015 m/s2 in a machine. The proton has an initial speed of 2.4 107 m/s and travels 2.8 cm.
(a) What is its speed?
(b) What is the increase in its kinetic energy?
a = 3.6*10^15 m/s^2
accelerating force = M*a = 6.01*10^-12 newtons
Final velocity = V2
Initial Velocity = V1 = 2.4*10^7 m/s
(a) M*V2^2/2 = M*V1^2/2 + F*X
Solve for V2
(b) KE increase = F*X
To find the answers to the given questions, we need to apply the equations of motion and the formulas for kinetic energy.
(a) To find the final speed of the proton, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (unknown)
- u is the initial velocity (2.4 x 10^7 m/s)
- a is the acceleration (3.6 x 10^15 m/s^2)
- s is the distance traveled (2.8 cm = 0.028 m)
Substituting the given values into the equation, we can solve for v:
v^2 = (2.4 x 10^7)^2 + 2(3.6 x 10^15)(0.028)
v^2 = 5.76 x 10^14 + 2.016 x 10^14
v^2 = 7.776 x 10^14
To find v, we take the square root of both sides:
v = √(7.776 x 10^14)
v ≈ 8.81 x 10^7 m/s
Therefore, the speed of the proton is approximately 8.81 x 10^7 m/s.
(b) The increase in kinetic energy is given by the formula:
ΔKE = (1/2)mv^2 - (1/2)mu^2
where:
- ΔKE is the increase in kinetic energy (unknown)
- m is the mass of the proton (1.67 x 10^-27 kg)
- v is the final velocity (8.81 x 10^7 m/s)
- u is the initial velocity (2.4 x 10^7 m/s)
Substituting the values into the formula:
ΔKE = (1/2)(1.67 x 10^-27)(8.81 x 10^7)^2 - (1/2)(1.67 x 10^-27)(2.4 x 10^7)^2
ΔKE ≈ 5.24 x 10^-14 Joules
Therefore, the increase in kinetic energy is approximately 5.24 x 10^-14 Joules.
To answer these questions, we can use the equations of motion and the work-energy theorem.
Step 1: Convert the given quantities to SI units:
Mass, m = 1.67 x 10^-27 kg
Acceleration, a = 3.6 x 10^15 m/s^2
Initial speed, u = 2.4 x 10^7 m/s
Distance, s = 2.8 cm = 2.8 x 10^-2 m
Step 2: Find the final speed, using the equation of motion:
v^2 = u^2 + 2as
Plugging in the given values, we have:
v^2 = (2.4 x 10^7 m/s)^2 + 2 * (3.6 x 10^15 m/s^2) * (2.8 x 10^-2 m)
v^2 = 5.76 x 10^14 m^2/s^2 + 2.016 x 10^14 m^2/s^2
v^2 = 7.776 x 10^14 m^2/s^2
Taking the square root of both sides, we get:
v ≈ 8.816 x 10^7 m/s
Therefore, the final speed of the proton is approximately 8.816 x 10^7 m/s.
Step 3: Find the increase in kinetic energy, using the work-energy theorem:
The increase in kinetic energy (ΔKE) of an object is equal to the net work done on it.
The net work done on the proton is given by:
Net work (W) = ΔKE
Using the equation for work, W = F * s, where F is the force applied and s is the displacement, we can find the net work done.
The force applied on the proton is given by:
F = m * a
Plugging in the given values, we have:
F = (1.67 x 10^-27 kg) * (3.6 x 10^15 m/s^2)
F = 6.012 x 10^-12 N
The displacement, s = 2.8 x 10^-2 m
Using the equation W = F * s, we can find the net work done:
W = (6.012 x 10^-12 N) * (2.8 x 10^-2 m)
W = 1.683 x 10^-13 J
Therefore, the increase in kinetic energy, ΔKE, is approximately 1.683 x 10^-13 J.