Answer:
a) The trajectory will be a helical path.
b) θ = 2*π rad
Explanation:
a) Since the initial velocity of the particle has a component parallel (x-component) to the magnetic field B
, the trajectory will be a helical path.
b) Given
t = 2*π*m/(q*B)
We can use the equation
θ = ω*Δt
where
θ is the angular displacement
ω is the angular speed, which is obtained as follows:
ω = q*B/m
then we have
θ = (q*B/m)*2*π*m/(q*B)
⇒ θ = 2*π rad
Answer:
Ball hit the tall building 50 m away below 10.20 m its original level
Explanation:
Horizontal speed = 20 cos40 = 15.32 m/s
Horizontal displacement = 50 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
50 = 15.32 t + 0.5 x 0 x t²
t = 3.26 s
Now we need to find how much vertical distance ball travels in 3.26 s.
Initial vertical speed = 20 sin40 = 12.86 m/s
Time = 3.26 s
Vertical acceleration = -9.81 m/s²
Substituting in s = ut + 0.5at²
s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²
s = -10.20 m
So ball hit the tall building 50 m away below 10.20 m its original level
Answer:
4 s
Explanation:
u = 19.6 m/s, g = 9.8 m /s^2
Let the time taken to reach the maximum height is t.
Use first equation of motion.
v = u + at
At maximum height, final velocity v is zero.
0 = 19.6 - 9.8 x t
t = 19.6 / 9.8 = 2 s
As the air resistance be negligible, is time taken to reach the ground is also 2 sec.
So, total time taken be the ball to reach at original point = 2 + 2 = 4 s
