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:
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.