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

measuring things 1 litre of water