How Far Will the Man Move When Colliding with the Woman on Frictionless Ice?

What happens when a 51-kg woman and an 82-kg man stand 12.0 m apart on frictionless ice and collide with each other?

Explanation:

Conservation of Momentum: The law of conservation of momentum states that the total momentum of an isolated system remains constant. In this scenario, the man and woman will have zero total momentum initially as they are at rest. When they collide, the total momentum will still be zero, but it will have been transferred between them.

Calculations:

Initially, both the man and woman have zero momentum:
pm,i = 0
pw,i = 0 After the collision, the total momentum is still zero:
pm,f + pw,f = 0 Using the conservation of momentum equation, we can solve for the final velocity of the man:
vm,f = -mw/mm * vw,f To find the distance the man has moved, we need to know the time taken for the collision. The average velocity can be calculated as:
vavg = (-mw/mm + 1)/2 * vw,f Solving for the time taken for the man to collide with the woman:
t = 12.0 m / [(-82 kg/51 kg + 1)/2 * 0 m/s]
t = 6.75 s By considering the absolute value of time, we find that the man does not move at all before colliding with the woman. Therefore, they were already 12.0 m apart and on frictionless ice.
← Microscopic reflections exploring angular magnification Conservation of momentum and kinetic friction how far do fred and brutus slide →

More Posts: