A82.2 kg ice skater, moving at 9.1 m/s, crashes into a stationary skater of equal mass. after the collision, the two skaters move as a unit at 4.55 m/s. suppose the average force a skater can experience without breaking a bone is 4472 n. if the impact time is 0.138 s, what is the magnitude of the average force each skater experiences?

F= -2710 N

Thus, magnitude of average force experienced by moving skater and by stationary skater is 2710 N  This is less than 4472 N, so the skaters will not break a bone.

Explanation:

Average acceleration of moving skater = (change in velocity) / (elapsed time).

 a= (4.55 - 9.1)/0.138.

a= -2275/69 m/s².

Using F = ma, average force experiences by the moving skater

F= 82.2 * (-2275/69 ).  

F= -2710 N

"By Newton's third law, the force experienced by the stationary skater at any given moment during the collision will be equal and opposite to that experienced by the moving skater. Therefore the magnitude of the average force experienced by the stationary skater will be the same as that experienced by the moving skater."

Thus, magnitude of average force experienced by moving skater and by stationary skater is 2710 N  This is less than 4472 N, so the skaters will not break a bone.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.