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el_diablo549
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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
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