Question

A car travelling at 12 m/s into a stationary truck of about 10 times the cars mass.

a. If the collision was completely inelastic, what velocity would the two travel at if the stuck together?

b. If the collision was completely elastic, what would be the velocities of the car and truck after the collision?

c. In order to exert a force of only 3500N on the truck during the collision, how much time would the collision have to take?

Answer 1

(A) Applying momentum conservation for the collision,

pi = pf

(m x 12) + (10m x 0) = (m + 10m) v

v = 1.09 m/s

(B) for elastic collision,

(12 x m) + 0 = (m v1) + (10m v2)

v1 + 10 v2 = 12

v1 = 12 - 10 v2

for elastic collision,

velocity of approach = velocity of separation

12 = v2 - v1 = v2 - 12 + 10 v2

v2 = 2.18 m/s

and v1 = 12 - 10v2 = -9.82 m/s

Speed of car = 9.8 m/s in opposite direction
Speed of truck = 2.2 m/s in original direction of car

(C) Impulse = change in momentum = F t

22m = 3500 t

t = 0.0063 m

put mass of car (m) in kg and get time in sec.