Respuesta :
Answer:
F1 = -2810.87N
F2 = 2810.87N
Their bones are safe.
Explanation:
Change in momentum = impulse
∆p = F∆t
F= ∆p/∆t
Given;
m1 = m2 = 75.2kg
v1i= 7.7m/s
v2i= 0m/s
v1f=v2f = 3.85m/s
∆t = 0.103s
Where;
m1 and m2 are mass of the skaters
v1i and v2i are the initial velocity of the skaters
v1f and v2f are the final velocity of the skaters
F1 and F2 are the forces experienced by skater 1 and skater 2
For skater 1
F1 = ∆p1/∆t
F1 = m1(v1f - v1i)/∆t
F1 = 75.2kg (3.85-7.7)/0.103
F1 = -2810.87N
For skater 2
F2 = ∆p2/∆t
F2 = m2(v2f - v2i)/∆t
F2 = 75.2kg (3.85-0)/0.103
F2 = 2810.87N
Therefore, their bone is safe since their average force is less than 4824N
Answer:
2810.8 N, no bone will break
Explanation:
We have,
m1=m2=75.2kg
vi1=7.7m/s
vi2=0m/s
vf1=vf2=3.85m/s
Fmax = 4824N
Favg=?
To calculate the average force we need to calculate the impulse I
According to the Impulse-Momentum Theorem
I=Δp=FavgΔt
We know that momentum is always conserved so Δp1=Δp2
As average force exerted on both the skaters will be same due to action-reaction phenomena we find impact of force for the second skater.
I=Δp2=Favg2Δt
mvf2-mvi2=Favg2Δt
∵vi=0 ∴mvi2=0
Favg2=mvf2/Δt
Favg2=(75.2×3.85)/0.103
Favg2=2810.8N
∵Favg2<Fmax⇒2810.8N<4824N
∴No bone will break
