A proton (mass m = 1.67 x 10:27 kg) is being accelerated along a straight line at 4.60 x 1010 m/s² in a machine. If the proton has an initial speed of 4.00 x 10 m/s and travels 1.10 cm, what then is (a) its speed and (b) the increase in its kinetic energy?
(a) Number
Units
(b) Number
Units
The Correct Answer and Explanation is :
We can solve the problem using basic kinematic and work-energy principles. Here’s a step-by-step solution:
Given:
- Mass of proton (m) = 1.67 × 10⁻²⁷ kg
- Acceleration (a) = 4.60 × 10¹⁰ m/s²
- Initial speed (v₀) = 4.00 × 10⁰ m/s
- Distance traveled (d) = 1.10 cm = 0.011 m
Part (a): Finding Final Speed (v)
We will use the kinematic equation that relates the initial velocity, final velocity, acceleration, and distance traveled:
[
v^2 = v_0^2 + 2ad
]
Where:
- ( v ) is the final speed
- ( v_0 ) is the initial speed
- ( a ) is the acceleration
- ( d ) is the distance traveled
Substituting the known values:
[
v^2 = (4.00 \times 10^0)^2 + 2 \times (4.60 \times 10^{10}) \times (0.011)
]
[
v^2 = 16 + 1.012 \times 10^{10}
]
[
v^2 = 1.012 \times 10^{10} + 16
]
[
v^2 \approx 1.012 \times 10^{10} \, \text{m}^2/\text{s}^2
]
Now, taking the square root:
[
v \approx \sqrt{1.012 \times 10^{10}} \, \text{m/s}
]
[
v \approx 1.006 \times 10^5 \, \text{m/s}
]
So, the final speed is approximately:
[
v \approx 1.01 \times 10^5 \, \text{m/s}
]
Part (b): Increase in Kinetic Energy (ΔK)
The increase in kinetic energy is given by:
[
\Delta K = \frac{1}{2} m (v^2 – v_0^2)
]
Substituting the known values:
[
\Delta K = \frac{1}{2} \times (1.67 \times 10^{-27}) \times \left( (1.01 \times 10^5)^2 – (4.00 \times 10^0)^2 \right)
]
First, calculate the squares:
[
(1.01 \times 10^5)^2 = 1.0201 \times 10^{10}
]
[
(4.00 \times 10^0)^2 = 16
]
Now, substitute:
[
\Delta K = \frac{1}{2} \times (1.67 \times 10^{-27}) \times (1.0201 \times 10^{10} – 16)
]
[
\Delta K = \frac{1}{2} \times (1.67 \times 10^{-27}) \times 1.0201 \times 10^{10}
]
[
\Delta K = 8.53 \times 10^{-18} \, \text{J}
]
So, the increase in kinetic energy is:
[
\Delta K \approx 8.53 \times 10^{-18} \, \text{J}
]
Final Answer:
(a) Speed = 1.01 × 10⁵ m/s
(b) Increase in Kinetic Energy = 8.53 × 10⁻¹⁸ J
Explanation:
- Part (a) calculates the final speed using kinematic equations, which rely on the relationship between acceleration, initial velocity, and distance traveled.
- Part (b) uses the work-energy theorem, which tells us that the work done on an object (in this case, the proton) is equal to the change in its kinetic energy. The kinetic energy increases as the proton accelerates, which is computed by subtracting the initial kinetic energy from the final kinetic energy.