1. a car carrying a 75 kg test dummy crashes into a wall at 25m/s and is brought to rest in 0.1s. Show that the average forced exerted by the seat belt on the dummy is 18,750N.
2. Judy (mass 40kg) standing on slippery ice, catches her leaping dog (mass 15kg) moving horazontally at 3.0 m/s. Show that the speed of Judy and her dog after the catch is 0.8m/s
3 Answers
for the first question, use a simple force equation.
F = ma. classic newton.
however, we dont actually have an acceleration here, so because a = change in velocity/change in time, we can plug in v/t for a.
F = mv/t.
plug in numbers:
F = (75kg)(25m/s-0m/s)/(0.1s). notice how i did the change in velocity (final minus initial)
F = 18750 N
for the second question, it is simply conservation of momentum. p(initial)=p(final).
m1v1=m2v2
judy is standing still and her dog is moving at 3 m/s.
so for m1v1, you have (40kg)(0m/s)+(15kg)(3m/s).
at the end, you combine their masses and they have the same final speed.
(40kg)(0m/s)+(15kg)(3m/s) = (55kg)(v(final)).
v(final) = 0.8 m/s
J = F x (time)
J = change in momentom
J initial = 75 x 25 = 1875
J final = 0
therefore: 1875= (.1)(F)
F = 18750
2. Momentum of dog initially: 15 x 3 = 45
Momentum of system equals initial momentum: 45 = (15 + 40) x (v)
v = .8 m/s
m1v1 = m2v2