The discussion revolves around an inelastic collision problem involving two cars with specified masses and initial directions. The cars lock together and slide a certain distance at a given acceleration, raising questions about momentum and the determination of fault in the collision.
Some participants have offered insights into the types of equations that may be relevant, such as conservation of momentum and kinematic equations. There is an ongoing exploration of the details provided in the problem, with participants questioning the implications of the motion after the collision and the meaning of certain terms.
Participants note the lack of initial velocity information and express confusion regarding the direction of acceleration and its relation to the motion of the combined mass. The problem's constraints and the need for further clarification on certain terms are acknowledged.
Just invent variables to represent the three unknown velocities, the two before and the one after collision.MomentumIsHard said:Homework Statement
Car 1: 2200kg
Initial direction: South
Car 2: 1800kg
Initial direction: East
Cars lock together and slide 11.25M (S 22.25 E)
At an acceleration that is 9.81 m (back) (it's a coincidence that this is gravity.)
Homework Equations
Who's fault was the collision if the speed limit was 50km/h?
I can't find momentum without a velocity so I'm very lost
The Attempt at a Solution
haruspex said:Just invent variables to represent the three unknown velocities, the two before and the one after collision.
Momentum gives you two equations, stopping distance gives you a third.
All the question states is that it moves 11.25 M [Back] (even though other directions were given in terms of east and south) at an acceleration of 9.81 m/s. The problem is I don't know where to start so I can't really make an attempt. Once I get one momentum I could pretty easily figure this out but I can't quite figure that out with no velocities givengneill said:Hi MomentumIsHard,
You need to make some attempt before help can be given. What details about the motion after the collision do you know?
Conservation of momentum (one for each of two dimensions) and the SUVAT equation relating speeds, distance and (constant) acceleration.MomentumIsHard said:Which equations are those?
No idea what "[Back]" is supposed to mean, but it doesn't matter if you're given the rest of the motion details. You have a distance, an acceleration, and a final speed. What can you deduce from that?MomentumIsHard said:All the question states is that it moves 11.25 M [Back] (even though other directions were given in terms of east and south) at an acceleration of 9.81 m/s. The problem is I don't know where to start so I can't really make an attempt. Once I get one momentum I could pretty easily figure this out but I can't quite figure that out with no velocities given
I think it means the direction of the given acceleration is opposite to the motion.gneill said:No idea what "[Back]" is supposed to mean, but it doesn't matter if you're given the rest of the motion details. You have a distance, an acceleration, and a final speed. What can you deduce from that?
That's quite likely. Of course deceleration is also implied by the fact that the combined mass travels a limited distance.haruspex said:I think it means the direction of the given acceleration is opposite to the motion.