A particle (q-4.0 mC.m-50 g) has a velocity of 25 m/s in the positive x direction when it first enters a region where the electric field is uniform (60 N/C in the positive y direction). What is the speed of the particle 5.0 s after it enters this region?
is simple steps
HINT: the answer is 35m\s
To solve this problem, we can use the equation:
F = qE
Where F is the force experienced by the particle, q is its charge, and E is the electric field. Since the force is in the y-direction, we can rewrite the equation as:
Fy = qEy
Using Newton's second law, we know that force equals mass times acceleration:
F = ma
Therefore, we can rewrite the equation as:
ma = qEy
Since we know the mass of the particle is 50 g (or 0.05 kg), the charge is 4.0 mC (or 4.0 * 10^-3 C), and the electric field is 60 N/C, we can substitute these values into the equation:
0.05a = (4.0 * 10^-3)(60)
Simplifying the right side of the equation:
0.05a = 0.24
Now, we can solve for acceleration:
a = 0.24 / 0.05 = 4.8 m/s^2
Acceleration is the rate of change of velocity over time, so we can use the equation:
a = (vf - vi) / t
Rearranging the equation to solve for the final velocity:
vf = a * t + vi
Plugging in the given values:
vf = (4.8)(5.0) + 25
vf = 24 + 25
vf = 49 m/s
However, the question asks for the speed of the particle, which is the magnitude of its velocity. In this case, since the particle is moving in the positive x direction, the velocity in the y direction does not contribute to the speed. Therefore, the speed is simply the magnitude of the x component of the velocity, which is 25 m/s.
To find the speed of the particle 5.0 s after it enters the region, you can use the principles of electrostatic forces and kinematics.
Step 1: Find the force acting on the particle due to the electric field.
The force(F) acting on a charged particle in an electric field(E) is given by the equation F = qE, where q is the charge of the particle. In this case, q = -4.0 mC and E = 60 N/C.
F = (-4.0 mC)(60 N/C)
F = -240 mN
Step 2: Use the force and mass to find the acceleration of the particle.
The Newton's second law of motion states that force(F) = mass(m) * acceleration(a). Rearranging the equation, we can find the acceleration.
F = ma
-240 mN = (50 g)(a) [Since mass is given as 50 g and acceleration is in the same direction as the force]
-0.240 N = (0.050 kg)(a) [1 N = 1 kg * m/s^2]
a = -4.8 m/s^2
Step 3: Find the change in speed using kinematic equations.
We have the initial velocity(u) of the particle as 25 m/s in the positive x direction. Let's assume the final velocity(v) after 5.0 s is v.
Using the formula v = u + at, where t = 5.0 s and a = -4.8 m/s^2:
v = 25 m/s + (-4.8 m/s^2)(5.0 s)
v = 25 m/s - 24 m/s
v = 1 m/s
Step 4: Find the final speed of the particle.
To find the magnitude of the velocity, we need to consider both the x and y components. The final speed can be found using the Pythagorean theorem:
Final speed = √(vx^2 + vy^2)
The x-component of velocity (vx) remains constant at 25 m/s, while the y-component of velocity (vy) can be calculated using the electric field and time elapsed.
vy = (60 N/C)(5.0 s)
vy = 300 m/s
Final speed = √(25 m/s)^2 + (300 m/s)^2)
Final speed = √(625 + 90000)
Final speed = √90625
Final speed ≈ 301.04 m/s
Therefore, the speed of the particle 5.0 s after entering the region is approximately 301.04 m/s.
To find the speed of the particle 5.0 seconds after it enters the region, we need to consider the force applied on the particle in the electric field and then calculate the resulting acceleration and velocity. Here are the steps to solve this problem:
Step 1: Identify the given values:
- Charge of the particle (q) = -4.0 mC
- Mass of the particle (m) = -50 g
- Initial velocity of the particle (v) = 25 m/s
- Electric field (E) = 60 N/C
- Time (t) = 5.0 s
Step 2: Identify the formula that relates force (F), charge (q), and electric field (E):
In this case, the formula is: F = q * E
Step 3: Calculate the electric force acting on the particle:
Using the formula F = q * E, we can substitute the given values:
F = ( -4.0 * 10^-3 C ) * ( 60 N/C )
F = -240 * 10^-3 N
F = -0.24 N
Step 4: Calculate the acceleration of the particle:
Since the force applied on the particle is constant, we can use Newton's second law:
F = m * a, where m is the mass of the particle
Rearranging the formula, we get:
a = F / m
Substituting the values:
a = ( -0.24 N ) / ( -50 * 10^-3 kg )
a = 4.8 m/s^2
Step 5: Calculate the final velocity of the particle:
Using the formula for uniformly accelerated motion:
v = u + a * t, where u is the initial velocity and t is the time
Substituting the values:
v = 25 m/s + ( 4.8 m/s^2 ) * ( 5.0 s )
v = 25 + 24
v = 49 m/s
Step 6: Calculate the speed of the particle:
The speed is the magnitude of velocity, so we can ignore any sign:
speed = |v|
speed = |49 m/s|
speed = 49 m/s
Therefore, the speed of the particle 5.0 seconds after entering the region is 49 m/s, not 35 m/s as specified in the hint.