A ball of 250g hits the floor at a velocity of 2,50 m/s at an angle of 70* relative to the vertical. The vertical force in function with time between the floor and the ball is:
from 0 to 50 N : from 0 to 1 sec.
from 50 N to 100 N : from 1 to 2 sec.
constant 100 N : from 2 to 3 sec.
100 to 50 N : from 3 to 4 sec.
and 50 N to 0 N : from 4 to 5 sec.
What is the velocity of the ball after the collision if the ball bounces? Give the answer in terms of unit vectors and polar notation.
To find the velocity of the ball after the collision, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Mass of the ball (m) = 250g = 0.25kg
Initial velocity (v) = 2.50 m/s
Angle of the velocity relative to the vertical (θ) = 70 degrees
To find the horizontal component (v_h) of the initial velocity:
v_h = v * cos(θ)
v_h = 2.50 m/s * cos(70°)
To find the vertical component (v_v) of the initial velocity:
v_v = v * sin(θ)
v_v = 2.50 m/s * sin(70°)
After the collision, the vertical component of the velocity will undergo a change in direction due to the bounce. However, the horizontal component will remain unchanged.
The vertical component can be found by analyzing the force acting on the ball during each time interval:
From 0 to 1 second (0 to 50 N):
The vertical force acting on the ball is increasing linearly from 0 to 50 N. This will result in a corresponding increase in the vertical velocity.
From 1 to 2 seconds (50 to 100 N):
The vertical force acting on the ball is constant at 100 N. This will result in a constant vertical velocity.
From 2 to 3 seconds (constant 100 N):
The vertical force acting on the ball remains constant at 100 N. The vertical velocity will continue to remain constant during this time interval.
From 3 to 4 seconds (100 to 50 N):
The vertical force acting on the ball decreases linearly from 100 N to 50 N. This will result in a corresponding decrease in the vertical velocity.
From 4 to 5 seconds (50 to 0 N):
The vertical force acting on the ball decreases linearly from 50 N to 0 N. This will result in a corresponding decrease in the vertical velocity until it comes to a halt.
To find the final vertical velocity (v_vf), we need to calculate the change in velocity during each time interval and then sum them up.
Change in velocity from 0 to 1 second:
Δv_v1 = (50 N - 0 N) * (1 sec - 0 sec) / 0.25 kg
Δv_v1 = 50 N * 1 sec / 0.25 kg
Change in velocity from 1 to 2 seconds:
Δv_v2 = (100 N - 50 N) * (2 sec - 1 sec) / 0.25 kg
Δv_v2 = 50 N * 1 sec / 0.25 kg
Change in velocity from 3 to 4 seconds:
Δv_v3 = (50 N - 100 N) * (4 sec - 3 sec) / 0.25 kg
Δv_v3 = -50 N * 1 sec / 0.25 kg
Change in velocity from 4 to 5 seconds:
Δv_v4 = (0 N - 50 N) * (5 sec - 4 sec) / 0.25 kg
Δv_v4 = -50 N * 1 sec / 0.25 kg
Now, we can find the final vertical velocity:
v_vf = v_v + Δv_v1 + Δv_v2 + Δv_v3 + Δv_v4
Finally, to find the final velocity of the ball (v_f), we combine the horizontal and vertical components:
v_f = sqrt(v_h^2 + v_vf^2)
The result will be in terms of unit vectors in polar notation, with the magnitude of the velocity and the direction angle.