A charged particle of mass m of 1 kg and charge q = 2 uC is thrown from a horizontal ground at an angle teta = 45 degree with speed 20 m/s. In space a horizontal electric field E = 2 x 10^7 V/m exist. Find the range on horizontal ground of the projectile thrown
v(ox) =v(o)•cosθ =20•0.707=14.14 m/s,
v(oy) =v(o)•sinθ =20•0.707=14.14 m/s.
The motion from the initial point to the highest point is accelerated in x-direction and decelerated in y-direction:
x: x=v(ox)•t+a•t^2/2
v(x) = v(ox) +a•t
y: y=v(oy)•t - g•t^2/2
v(y) = v(oy) - g•t.
For the highest point
v(y) = 0, t = v(oy)/g = v(o)•sinθ/g = 20•0.707/9.8 = 1.443 s.
ma = qE a = qE/m = 2•10^-6•2•10^7/1 = 40 m/s^2
L1 = v(ox)•t+a•t^2/2 =14.14•1.443 +40•(1.443)^2/2 =62.04 m.
v(x) = v(ox) +a•t =14.14 + 20•1.443 = 43 m/s.
This speed is the initial speed for the second part of horizontal motion
V(o) = 43 m/s.
The motions in vertical and horizontal directions are independent from each other, therefore, for vertical motion the time of descent is t =1.443 s.
For the motion from the highest point to the final point
L2= V(o)•t+a•t^2/2 =43•1.443 + 40•(1.443)^2/2 =103.7 m.
L = L1 +L2 = 62.04 + 103.7 = 165.74 m.
To find the range on the horizontal ground of the projectile, we need to consider both the horizontal velocity and the time of flight of the particle.
First, let's find the horizontal velocity component of the projectile. The initial velocity of the particle can be split into horizontal and vertical components as follows:
Vx = V * cos(theta)
Vy = V * sin(theta)
Given that the speed, V, is 20 m/s and the angle, theta, is 45 degrees, we can calculate:
Vx = 20 m/s * cos(45 degrees)
Vx = 20 m/s * sqrt(2)/2
Vx = 10 * sqrt(2) m/s
The horizontal velocity component, Vx, is 10 * sqrt(2) m/s.
Next, let's find the time of flight of the particle. To do this, we need to find the vertical component of the velocity, which is affected by the electric field. The electric field provides an upward force on the particle, which reduces the acceleration due to gravity.
The net force on the particle in the vertical direction is:
F_net = q * E - m * g
where q is the charge of the particle, E is the electric field strength, and g is the acceleration due to gravity.
Given that q = 2 uC = 2 * 10^-6 C, E = 2 * 10^7 V/m, and g = 9.8 m/s^2, we can calculate the net force:
F_net = (2 * 10^-6 C) * (2 * 10^7 V/m) - (1 kg) * (9.8 m/s^2)
F_net = (4 * 10^-6 kg*m/s^2) - (9.8 kg*m/s^2)
F_net = -5.8 kg*m/s^2
The negative sign means that the net force is acting in the downward direction, opposing the force of gravity.
Now we can use Newton's second law, F = m * a, to find the acceleration in the vertical direction:
F_net = m * a
-5.8 kg*m/s^2 = (1 kg) * a
a = -5.8 m/s^2
Again, the negative sign indicates that the acceleration is directed downward.
Since the electric field opposes the force of gravity, the net acceleration is the difference between the acceleration due to gravity and the electric field. Therefore, the acceleration due to gravity is reduced by 5.8 m/s^2.
Next, let's find the time of flight. Since the acceleration is in the vertical direction, we can use the following equation of motion:
Vy = (Vy0) + a * t
Given that the initial vertical velocity, Vy0, is 20 m/s * sin(45 degrees) = 10 * sqrt(2) m/s, and the final vertical velocity, Vy, is 0 m/s (at the top of the projectile's trajectory), we can calculate the time of flight:
0 = (10 * sqrt(2) m/s) + (-5.8 m/s^2) * t
10 * sqrt(2) m/s = 5.8 m/s^2 * t
t = (10 * sqrt(2) m/s) / (5.8 m/s^2)
t ≈ 1.76 s
The time of flight, t, is approximately 1.76 seconds.
Finally, to find the range, we multiply the horizontal velocity, Vx, by the time of flight, t:
Range = Vx * t
Range = (10 * sqrt(2) m/s) * (1.76 s)
Range ≈ 17.6 m
Therefore, the range on the horizontal ground of the projectile thrown is approximately 17.6 meters.