A man claims that he can hold onto a 16.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 59.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.05 s. Find the magnitude of the average force needed to hold onto the child.
N

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

Explanation:

Given;

Mass of child m = 16 kg

Speed of each car v = 59.0 mi/h = 26.37536 m/s

Time t = 0.05s

Applying the impulse momentum equation;

Impulse = change in momentum

Ft = ∆(mv)

F = ∆(mv)/t

F = m(∆v)/t

Where;

F = force

t = time

m = mass

v = velocity

Since the final speed of the car is zero(at rest) then;

∆v = 0 - v = -26.37536 m/s

Substituting the given values;

F = 16×-26.37536/0.05

F = -8440.1152 N

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

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.