A 2250 kg truck is pulling out of a drive way. A 1500 kg sedan strikes it on the left hand side. Moves the 2250 kg 4×4 truck 10ft to the right and spins it 90 degrees. My question is how much force is needed and what the speed of the sedan going into the collision
?
Did they really give you Feet?
10 ft = 3.05 meters
sedan was going at speed u
so original linear momentum was 1500 u
after the collision assuming they stick together the momentum is
(1500 + 2250) v = 3750 v
momentum after = momentum before so
1500 u = 3750 v
now they stop together going from speed v to 0 in 3.05 meters
average speed = 0.5 v
time = distance/average speed = 3.05/ 0.5 v
change in momentum = 3750 v
force = change in momentum / time = 3750 v /( 3.05 / 0.5 v) = 615 v^2
we need to know one more thing to finish., like coef of friction or something