Carousel dynamics question involving collision need to check one part

[el_diablo549]

Homework Statement

find the velocity that the man was traveling at immediately before the collision
radius = 10m
height = 3m
angular velocity = constant
mass of person = 65kg
mass of kiosk = 150kg
displacement of kiosk after collision = 5m
coefficient of friction kiosk = 0.6

Homework Equations

f = ma
frictional force = mass x gravity x friction co-efficient
suvat equations
Mi1Vi1 + Mi2Vi2 = Mf1Vf1 + Mf2Vf2 conservation of momentum

The Attempt at a Solution

combine both mass so m = 215kg
multiply by 0.6 for friction and -9.81 m/s^2 for gravity
total frictional force = -1265.49N
if F = Ma then a = F/M so acceleration is -5.886 m/s^2
we now have
a = -5.886m/s^2
s = 5m
v(final velocity) = 0 m/s
t = irrelevant
u(initial velocity) variable we need to find

using v^2 = u^2 + 2as

we find u = 7.672 m/s
which is the initial velocity of the combine mass of the guy and the kiosk immediately after the collision as they travel together for a distance of 5m

using
Mi1Vi1 + Mi2Vi2 = Mf1Vf1 + Mf2Vf2

we find the velocity that the man was traveling at immediately before the collision is 25.3768 m/s

Is this correct? the problem I see with it is if you draw the free body diagram it shows the frictional force is -1265.49N and I am using the force of the collision as 1265.49N which would mean the sum of forces in x direction = 0 and the object is in equilibrium and not moving at all which contradicts my whole answer

 

[PhanthomJay]

View attachment 82365

Homework Statement

find the velocity that the man was traveling at immediately before the collision
radius = 10m
height = 3m
angular velocity = constant
mass of person = 65kg
mass of kiosk = 150kg
displacement of kiosk after collision = 5m
coefficient of friction kiosk = 0.6

Homework Equations

f = ma
frictional force = mass x gravity x friction co-efficient
suvat equations
Mi1Vi1 + Mi2Vi2 = Mf1Vf1 + Mf2Vf2 conservation of momentum

The Attempt at a Solution

combine both mass so m = 215kg
multiply by 0.6 for friction and -9.81 m/s^2 for gravity
total frictional force = -1265.49N
if F = Ma then a = F/M so acceleration is -5.886 m/s^2
we now have
a = -5.886m/s^2
s = 5m
v(final velocity) = 0 m/s
t = irrelevant
u(initial velocity) variable we need to find

using v^2 = u^2 + 2as

we find u = 7.672 m/s
which is the initial velocity of the combine mass of the guy and the kiosk immediately after the collision as they travel together for a distance of 5m

using
Mi1Vi1 + Mi2Vi2 = Mf1Vf1 + Mf2Vf2

we find the velocity that the man was traveling at immediately before the collision is 25.3768 m/s

Is this correct? the problem I see with it is if you draw the free body diagram it shows the frictional force is -1265.49N and I am using the force of the collision as 1265.49N which would mean the sum of forces in x direction = 0 and the object is in equilibrium and not moving at all which contradicts my whole answer

Your answer seems correct. The collision force happens impulsively over a short period of time and is not determinable with the information given, but it is not the same as the friction force and it is not needed to solve this problem. The only net force acting on the man-kiosk system after the impact is the net friction force .

 

[el_diablo549]

Your answer seems correct. The collision force happens impulsively over a short period of time and is not determinable with the information given, but it is not the same as the friction force and it is not needed to solve this problem. The only net force acting on the man-kiosk system after the impact is the net friction force .

hey thanks man that makes sense so the only forces acting on it immediately after the collision is gravity, normal force and the frictional force which means it has an acceleration of - 5.886m/s^2 over 5m which determines its initial velocity if its final velocity is 0

 

[PhanthomJay]

hey thanks man that makes sense so the only forces acting on it immediately after the collision is gravity, normal force and the frictional force which means it has an acceleration of - 5.886m/s^2 over 5m which determines its initial velocity if its final velocity is 0

Precisely. Your solution is very well presented, excellent work.

 

[el_diablo549]

Precisely. Your solution is very well presented, excellent work.

Am I also correct in assuming when he is launched from the carousel, the path he travels in is a straight line perpendicular to the radius meaning the distance the kiosk is from the base shaft is the square root of the radius squared plus (the 5 metres the kiosk traveled plus the distance the man flew from the carousel) squared?

 

[PhanthomJay]

Am I also correct in assuming when he is launched from the carousel, the path he travels in is a straight line perpendicular to the radius meaning the distance the kiosk is from the base shaft is the square root of the radius squared plus (the 5 metres the kiosk traveled plus the distance the man flew from the carousel) squared?

The problem asks for the distance from the kiosk to the (base of) the shaft before the accident, so don't add in that 5 meters that the kiosk traveled after the accident. Otherwise, yes, the man travels tangent to the curve in a parabolic projectile path to the kiosk, so you have to determine the horizontal distance from the kiosk in its original position to the man at the moment he leaves the carousel, then take the sq rt of the radius squared plus that distance squared.