A75.2 kg ice skater, moving at 7.7 m/s, crashes into a stationary skater of equal mass. after the collision, the two skaters move as a unit at 3.85 m/s. suppose the average force a skater can experience without breaking a bone is 4824 n. if the impact time is 0.103 s, what is the magnitude of the average force each skater experiences?

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

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

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