2.Two cars with the same mass,and traveling at the same spee

I would go with C.P= F * t,the second car is completely crumpled its front,indicating that the front end deformed during the collision,thus slowed the speed of the car.In other word,a lot of the momentum was absorbed by its front end.
the impulse momentum theorem:the change in momentum P equals to the impulse of the force F.I = change in P ==> I = F*t,the average force during the deformation of the second car,times,the time duration of the collision caused deformation t,equals to the momentum decreased by the second car.
So,since the momentum of the car is decreased,the people in the car is less likely to be injured.However,in the first case,the front of car is so rigid that it did not damage at all,all the momentum is then subjected to decrease to zero (car to stop) in very short period of time t.momentum change = F * t,then since the time is short,the average force (to make people stop) of the people experienced F will be much higher,in other word,it is then more dangerous.

。。。。

两辆一样的车对撞，由于初速度跟开速度的大小相同，也就是说变化的冲量相等，但变形小的汽车缓冲时间更小，因此造成的冲击力更大，所以答案为C

你是老外吗？不对，你应该是很想请个外教。