a. impulse=changemomentum, you are given impulse, solve for change of momentum
b. force*time=impulse
avg force=impulse/time
c. avg force= impulse/(3*time)
In a crash test, a 2000-kg van collides with a concrete support. The stopping time interval for the collision is 0.10 s, and the impulse exerted by the support on the van is 7.0×103N⋅s.
A.) Determine the magnitude of the change in momentum.
B.)Determine the magnitude of the average force exerted by the concrete support on the van.
C.)If the van is constructed to collapse more during the collision so that the time interval during which the impulse is exerted is tripled, what is the magnitude of the average force exerted by the concrete support on the van?
2 answers
a) ANSWER: 5.5 * 10^3 N*s
impulse = change in momentum; I = 5.5 *10^3 so the change in momentum = 5.5 * 10^3
B) ANSWER 55000 N
impulse = Faverage * time
5.5 * 10^3 = Faverage * (0.1)
5.5 * 10^4 = Faverage
C) ANSWER 18000 N
impulse = Faverage * time
Our new time is 0.3 (since 0.1 * 3 = 0.3)
5.5 * 10^3 = Faverage * (0.3)
1.8 * 10^4 = Faverage (Faverage decreases)
impulse = change in momentum; I = 5.5 *10^3 so the change in momentum = 5.5 * 10^3
B) ANSWER 55000 N
impulse = Faverage * time
5.5 * 10^3 = Faverage * (0.1)
5.5 * 10^4 = Faverage
C) ANSWER 18000 N
impulse = Faverage * time
Our new time is 0.3 (since 0.1 * 3 = 0.3)
5.5 * 10^3 = Faverage * (0.3)
1.8 * 10^4 = Faverage (Faverage decreases)
1 answer