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cheff3r
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there are three parts to my problem (you can probably tell i haven't done much work on resitution before)
A bioengineer studying helmet design uses an experimental apparatus that launches a 2.4 kg helmet containing a 2 kg model of the human head against a rigid surface at 6 m/s. The head, suspended within the helmet, is not immediately affected by the impact of the helmet with the surface and continues to move to the right at 6 m/s, so the head then undergoes an impact with the helmet. If the coefficient of restitution of the helmet’s impact with the surface is 0.85 and the coefficient of restitution of the subsequent impact of the head with the helmet is 0.15
a) what is the velocity of the head after its initial impact with the helmet?
b) If the duration of the impact of the head with the helmet is 0.004 s, what is the magnitude of the average force exerted on the head by the impact?
c) Suppose that the simulated head alone strikes the rigid surface at 6 m/s, the coefficient of restitution is 0.5, and the duration of the impact is 0.0002 s. What is the magnitude of the average force exerted on the head by the impact?
So we know these formulas (possibly more that i didn't think of)
0.5*m1*v1^2+0.5*m*v2^2=0.5*m1*v3^2+0.5*m2*v4^2
F=ma
v3-v4=-e(v1-v2)
part a)
this is the part I'm struggling on
i was going to use v3-v4=-e(v1-v2) for the helmet and the wall to work out velocity of helmet after impact first but that is wrong since it has not taken into account of the head hitting the helmet
my solution part b)
F=ma which is equal to F=mv/t
F=2v/0.004 (v being answer to part a which i can't work out)
F=500v (is this the right method?)
my solution part c)
using equation v3-v4=-e(v1-v2) (coefficent of restitution formula)
v1=6m/s
v4=0m/s
v2=0m/s
e=0.5
v3=-0.5(6-0)+0
v3=-3
v3=3 m/s in oppostie direction (is this right?)
F=ma which is equal to F=mv/t
F=2*3/0.0002
F30,000N
A bioengineer studying helmet design uses an experimental apparatus that launches a 2.4 kg helmet containing a 2 kg model of the human head against a rigid surface at 6 m/s. The head, suspended within the helmet, is not immediately affected by the impact of the helmet with the surface and continues to move to the right at 6 m/s, so the head then undergoes an impact with the helmet. If the coefficient of restitution of the helmet’s impact with the surface is 0.85 and the coefficient of restitution of the subsequent impact of the head with the helmet is 0.15
a) what is the velocity of the head after its initial impact with the helmet?
b) If the duration of the impact of the head with the helmet is 0.004 s, what is the magnitude of the average force exerted on the head by the impact?
c) Suppose that the simulated head alone strikes the rigid surface at 6 m/s, the coefficient of restitution is 0.5, and the duration of the impact is 0.0002 s. What is the magnitude of the average force exerted on the head by the impact?
So we know these formulas (possibly more that i didn't think of)
0.5*m1*v1^2+0.5*m*v2^2=0.5*m1*v3^2+0.5*m2*v4^2
F=ma
v3-v4=-e(v1-v2)
part a)
this is the part I'm struggling on
i was going to use v3-v4=-e(v1-v2) for the helmet and the wall to work out velocity of helmet after impact first but that is wrong since it has not taken into account of the head hitting the helmet
my solution part b)
F=ma which is equal to F=mv/t
F=2v/0.004 (v being answer to part a which i can't work out)
F=500v (is this the right method?)
my solution part c)
using equation v3-v4=-e(v1-v2) (coefficent of restitution formula)
v1=6m/s
v4=0m/s
v2=0m/s
e=0.5
v3=-0.5(6-0)+0
v3=-3
v3=3 m/s in oppostie direction (is this right?)
F=ma which is equal to F=mv/t
F=2*3/0.0002
F30,000N