What Speed Does the Second Car Achieve After a Collision?

[pokeefer]

vector momentum problem

Homework Statement

A car moving North at 9.0 m/s strikes a stationary car of equal mass. The first car moves off after the collision at an angle of 30 degrees West of North with a speed of 6.0 m/s.

(a) What is the speed of the second car after the collision? (3 marks)

(b) Show that the collision is inelastic. (2 marks)

(c) Explain how dents, skid marks, etc., show that kinetic energy has been lost (2 marks)

(d) What would the speeds of the cars be if the first car moved off at 30 degrees from a perfectly elastic collision? (3 marks)

Homework Equations

v1 + v2 = v1' + v2'

The Attempt at a Solution

I'm pretty sure I messed up on my calculation when I used the cosine law. The variable I'm trying to solve for is v2'

9 + 0 = 6 + v2'

Here was the calculation I made for part a:

v2'^2 = v1^2 - v1'^2 - 2(v1)(v1')(cos30)

Which I don't think is right.

I'm stumped at the other parts as well.

 

[srmeier]

I don't believe law of cosines will get you anywhere. Think about breaking your velocity vectors into their components.

 

[pokeefer]

Since I'm trying to solve for the velocity of a stationairy object after a collision. I'm adding vectors?

Also if the collision is inelastic how do I know what angle the second car moved off at?

 

[srmeier]

You're not "adding vectors." You need to apply conservation of momentum. Momentum is a vector quantity. Thus if you know the momentum vector before the collision, the sum of the individual momenta afterwards must equal what?

 

[rwisz]

Hello,

I can understand your confusion with this vector momentum problem. Let's go through each part step by step to try and find the correct solution.

a) To solve for the speed of the second car after the collision, we can use the conservation of momentum equation: m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the two cars, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities. We know that the mass of both cars is equal, so we can simplify the equation to: v1 + 0 = v1' + v2'. From this, we can solve for v2' by substituting the given values: 9 + 0 = 6 + v2', which gives us v2' = 3 m/s.

b) To show that the collision is inelastic, we can calculate the coefficient of restitution (e) using the formula: e = (v2' - v1') / (v1 - v2), where v1 and v2 are the initial velocities, and v1' and v2' are the final velocities. If the value of e is less than 1, then the collision is inelastic. In this case, we get e = (3 - 6) / (9 - 0) = -1/3, which is less than 1, therefore the collision is inelastic.

c) Dents, skid marks, and other physical evidence can show that kinetic energy has been lost in a collision. This is because in an inelastic collision, some of the initial kinetic energy is converted into other forms of energy, such as heat and sound, due to the deformation of the objects involved. This loss of kinetic energy can be seen in the damage to the cars and the marks left on the road.

d) In a perfectly elastic collision, kinetic energy is conserved, meaning that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. So, for part d, we can use the equation: 1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2, where m1 and m2 are the masses of the