A fugitive tries to hop on a freight train traveling at a constant speed of 4.8 . Just as an empty box car passes him, the fugitive starts from rest and accelerates at = 1.5 m/s^2 to his maximum speed of 5.9 m/s . a. How long does it take him to catch up to the empty box car? b. What is the distance traveled to reach the box car?
1.A proton, traveling with a velocity of 4.5 x106m/s due east,
experiences a magnetic force that has a maximum magnitude of
8.0 x 1014N and a direction of due south. What are the magnitude and
direction of the magnetic field causing the force? (b) Repeat part (a)
assuming the proton is replaced by an electron.
The Correct Answer and Explanation is :
Problem 1: Fugitive Catching Up to Box Car
Part A: How long does it take the fugitive to catch up to the empty box car?
To solve this, we need to consider the motion of both the fugitive and the box car.
- Fugitive’s Motion:
The fugitive starts from rest and accelerates at (a = 1.5 \, \text{m/s}^2) to a maximum speed of (v = 5.9 \, \text{m/s}). The equation for the time (t) taken to reach the maximum speed can be found using the equation: [
v = at
]
Solving for (t): [
t = \frac{v}{a} = \frac{5.9}{1.5} = 3.93 \, \text{seconds}
] So, it takes the fugitive 3.93 seconds to reach his maximum speed. - Box Car’s Motion:
The box car is traveling at a constant speed of (4.8 \, \text{m/s}).
Now, let’s find the distance the fugitive travels during the acceleration phase and the distance the box car travels in that time:
- Distance traveled by the fugitive during acceleration is: [
d = \frac{1}{2} a t^2 = \frac{1}{2} \times 1.5 \times (3.93)^2 = 11.6 \, \text{meters}
] - Distance traveled by the box car in the same time is: [
d_{\text{box car}} = v_{\text{box car}} \times t = 4.8 \times 3.93 = 18.864 \, \text{meters}
]
Now, the fugitive needs to catch up to the box car, so we can set up the equation for the time (t_{\text{catch}}) after the fugitive reaches his maximum speed. At that point, the fugitive travels at a constant speed of 5.9 m/s, and the box car continues at 4.8 m/s. The distance between them is:
[
\text{Initial distance between them} = d_{\text{box car}} – d = 18.864 – 11.6 = 7.264 \, \text{meters}
]
Let’s calculate the time taken for the fugitive to catch up to the box car after reaching maximum speed:
[
\text{Relative velocity} = 5.9 – 4.8 = 1.1 \, \text{m/s}
]
[
t_{\text{catch}} = \frac{\text{Distance}}{\text{Relative velocity}} = \frac{7.264}{1.1} = 6.6 \, \text{seconds}
]
Thus, the total time for the fugitive to catch up to the box car is:
[
t_{\text{total}} = 3.93 + 6.6 = 10.53 \, \text{seconds}
]
Part B: Distance Traveled to Reach the Box Car
Now, we need to find the distance the fugitive travels in the total time. The fugitive accelerates for 3.93 seconds, then moves at a constant speed for 6.6 seconds.
- Distance traveled during acceleration: [
d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 1.5 \times (3.93)^2 = 11.6 \, \text{meters}
] - Distance traveled at constant speed: [
d_2 = v \times t = 5.9 \times 6.6 = 38.94 \, \text{meters}
]
Thus, the total distance traveled by the fugitive is:
[
d_{\text{total}} = d_1 + d_2 = 11.6 + 38.94 = 50.54 \, \text{meters}
]
Problem 2: Magnetic Force on a Proton and Electron
(a) Magnetic Force on a Proton
We can use the equation for the magnetic force on a charged particle moving through a magnetic field:
[
F = qvB \sin(\theta)
]
where:
- (F = 8.0 \times 10^{-14} \, \text{N}) (the maximum force),
- (v = 4.5 \times 10^6 \, \text{m/s}) (velocity of the proton),
- (q = 1.6 \times 10^{-19} \, \text{C}) (charge of the proton),
- (\theta = 90^\circ) (since the direction of force is due south, perpendicular to the proton’s eastward motion).
Since the force is at maximum, (\sin(\theta) = 1), and the equation becomes:
[
F = qvB
]
Solving for (B):
[
B = \frac{F}{qv} = \frac{8.0 \times 10^{-14}}{(1.6 \times 10^{-19})(4.5 \times 10^6)} = 1.11 \, \text{T}
]
Thus, the magnitude of the magnetic field is (1.11 \, \text{T}), and its direction is due upward (perpendicular to both the velocity and the force, according to the right-hand rule).
(b) Magnetic Force on an Electron
When the proton is replaced by an electron, the charge of the electron is (-1.6 \times 10^{-19} \, \text{C}). However, the magnitude of the charge remains the same, so the magnetic field remains unchanged at (1.11 \, \text{T}).
However, the direction of the force changes due to the negative charge of the electron. Using the right-hand rule, we find that the force on the electron will be directed northward, opposite to the proton’s force direction.
Explanation
In both cases, the magnetic field causes the charged particles to experience a force that is perpendicular to both their velocity and the direction of the field. The magnitude of the force depends on the charge, velocity, and magnetic field strength, while the direction is determined using the right-hand rule (for positive charges) and the left-hand rule (for negative charges).
Image
I’ll now generate an image of the situation illustrating the forces, directions, and magnetic field for the proton and electron.
Here is the diagram illustrating the motion of the proton and electron in the magnetic field, showing their paths, force directions, and the magnetic field’s orientation. The proton’s force is directed south, while the electron’s force is directed north due to its negative charge.
