Average force and angle, collision problem

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AncientOne99
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Homework Statement


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Ball with a mass of 0.1 kg travels at a speed of 10 m /s to the boy.
Boy kicks the ball with foot so that it bounces back at an angle of 30 degrees with respect to the horizontal floor. The ball falls back to the ground 20 meters from the boy.
With what average force and under what angle with respect to the horizontal floor did boy kick the ball if the collision took 0.1 second?

Known data:
m = 0.1 kg
v0 = 10 m/s
β = 30°
Range = 20 m
Δt = 0.1 s

F = ?
α = ?

Homework Equations


p = m*v
FΔt = mΔv = Δp
Range = (v1^2*sin(2α))/g

The Attempt at a Solution


p0= m*v0
F =mΔv/Δt
Δv = v1(-x) - v0(x) ⇒ v1 + v0
v1=√(Range*g/(sin(2α))
tanβ = (v1y)/(v1x) ?
v1= √((v1x^2)+(v1y^2))

I cannot do anything with angle of refraction because i don't know velocity1(x) and velocity1(y) pls help, thank you.
 
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Probably i need to solve for α first. But because velocity 1 changes both x and y components because of average force i don't find a solution.
 
No α ≠ 30°, β = 30°
 
No α ≠ 0, but we are searching for α.
α= ?
 
AncientOne99 said:
No α ≠ 0, but we are searching for α.
α= ?

You are trying to find the angle that the force was applied. You need to do a vector diagram for force/momentum.

Note that the angle of the force is not necessarily the angle of the trajectory.
 
Please define all your variables.
AncientOne99 said:
v1=√(Range*g/(sin(2α))
According to that equation you are defining α as the angle the ball makes to the horizontal just after being kicked. That is the given angle of 30°. The unknown angle is that of the applied impulse.

The question does not give the flight angle of the ball before it is kicked. You will have to assume it is zero (say) or there is not enough information.
 
m = 0.1 kg
v0 = 10 m/s (x) direction
Refractive angle of ball = 30 ° (horizontal)
That's the angle of impact.
Range of ball = 20 m
time of impulse = 0.1 s

We are searching for:
1. Angle when the ball was kicked with respect to horizontal.
2. Average force of change in momentum.

Solution for problem:
Angle = 18 °
Average force = 24.2 N
 
I saw some trigonometric identity for this problem but i don't understand it geometrically, any idea?
240728
 
AncientOne99 said:
Refractive angle of ball = 30 ° (horizontal)
That's the angle of impact.
No it isn't. It is the angle at which the ball rises just after being kicked:
AncientOne99 said:
it bounces back at an angle of 30 degrees with respect to the horizontal floor
The angle at which the impulse was applied to the ball is to be found.

AncientOne99 said:
Angle when the ball was kicked with respect to horizontal.
It is unclear what that means. Angle of what when the ball was kicked?
 
With what average force and under what angle with respect to the horizontal floor did boy kick the ball if the collision took 0.1 second?
 
AncientOne99 said:
With what average force and under what angle with respect to the horizontal floor did boy kick the ball if the collision took 0.1 second?
Quite.
The ball is initially moving (horizontally at ground level, you will need to assume) at u=10m/s.
The boy's kick is an impulse of magnitude p at an angle to be determined (please say whether you want to call this α or β). The ball then rises at speed v and at an angle of θ=30 degrees to the horizontal and lands s=20m away.
From s and θ you can find v.

My recommendation is always to keep everything symbolic, only plugging in numbers at the end. There are numerous benefits.
 
Solution is incorrect for angle and force then
 
Yes, of course I am solving it analitically.
 
Ok, i will send you my thinking for this example.
 
Right, alfa is 30 degrees, i will just check if its right now
 
So we are searching horizontal angle of the FORCE
 
AncientOne99 said:
Right, alfa is 30 degrees, i will just check if its right now
This is very confusing. What are you meaning by alpha there? I asked you to decide whether to call the angle of the applied impulse alpha or beta. I used theta for the 30 degree angle the ball flies up at.
 
Ok, solved. Thank you very much for assistance. Mark