Two Skaters Collide (inelastic collision)

[ArielleDubois]

I'm trying to understand the concept of the following situation:

A 75.0-kg ice skater moving at 10.5 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 5.25 m/s. Suppose the average force a skater can experience without breaking a bone is 3,867 N. If the impact time is 0.112 s, does a bone break?

By using F= (MV)/t = [(75 kg) (5.25 m/s)] / 0.112s I found what I believe may be the correct answer (3515.63 N) so not a single bone breaks, but what I'm not understanding is the concept behind this solution.
Any help would be much appreciated.

 

[Liquidxlax]

Welcome to Physicsforums ArielleDubois. I'm not to sure where you got the equation of (MV)/t = F

but F = dp/dt

What you need is the equation for an inellastic collision which is

M₁v₁ + M₂v₂ = (M₁ + M₂)v₃

This represents a conservation of momentum. I mean obviously there is a force acting on both persons when they collide

You could use the change in momentum divided by the contact time to calculate the force I'm guessing since (dp/dt) is an infentesimal change in momentum w.r.t. a infinitesimal change in time

I'm not totally sure conisdering I've never saw a problem like this without using calculus.

I'd say to use

M₁v₁ + M₂v₂ = (M₁ + M₂)v₃

were v₂ = 0 since the second person is stationary.

I'm thinking

{M₁(v₃-v₁)}/t will get you the force for person in motion and

{Mv₃3}/t will get you the force for the stationary person

I'm not 100% sure maybe someone will be able to confirm this

 

[Rome_Leader]

To the above, MV/t appears to equal the Impulse, given that Momentum (p) = mass * velocity where M and V are velocity, and impulse is the change in momentum with respect to time.

The idea here is, as the person above has also said, to remember momentum is conserved for an inelastic collision. In the beginning, we have two separate entities, a skater moving, with a momentum equal to her mass times her velocity, and a stationary skater with no initial momentum.

As a result, the final momentum must be equal to the moving skater's momentum, since final momentum = initial momentum. Since the collision is perfectly inelastic (the once independent masses join as one), we can express final momentum as vf (m1 + m2).

Since you know m1, m2 and vf, you can get the final momentum! From there, use the impulse equation above! Your only error is you have forgotten to add the mass of the other skater! The rest is correct, but your answer will change.

Hope this helps!

 

[ArielleDubois]

Thank you for both of your replies! Here is the problem, as "Rome_Leader," noted in my calculations I only used one of the masses. Since the correct answer for this question is,"no, not a single bone breaks," then in my calculations I should yield a total force that is less than 3867 N, I only obtain such an answer when using one of the masses and not combining both m1 and m2, that is why I am confused. You are also correct, "Rome_Leader," I did use the formula for impulse but instead of finding the impulse I found the force, I used this formula because the question is interested in the resultant force of the collision.

 

[asteriatic]

I can explain the concept of this situation using the principles of momentum and impulse. In an inelastic collision, the two objects involved stick together and move as a single unit after the collision. This is different from an elastic collision where the objects bounce off each other.

In this situation, the two skaters have a combined mass of 150 kg and a combined initial momentum of (75 kg)(10.5 m/s) = 787.5 kg*m/s. After the collision, they have a combined momentum of (150 kg)(5.25 m/s) = 787.5 kg*m/s. This shows that momentum is conserved in the collision.

The average force experienced by an object can be calculated using the equation F = Δp/Δt, where Δp is the change in momentum and Δt is the time over which the change occurs. In this case, the change in momentum is 787.5 kg*m/s and the time is 0.112 s. This gives us an average force of 3515.63 N, which is less than the maximum force of 3867 N that a skater can experience without breaking a bone.

Therefore, based on these calculations, it is unlikely that a bone would break in this collision. However, it is important to note that these calculations are based on idealized conditions and do not take into account factors such as the angle of impact, the strength of the bones, and the distribution of force on the body. In a real-life scenario, there may be a greater risk of injury.