Two bumper cars crash head on in an elastic collision. One car #1 was moving 2.5 m/s west and car #2 was moving 3.7 m/s east. Each car has a mass of 254 kg. What are the final velocities after the collision?
I think this problem is supposed to be solved with m1v1, i + m2v2, i = m1v1, f + m2v2, f
Yes, you are correct. The problem involves an elastic collision, and according to the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. The equation you mentioned, m1v1i + m2v2i = m1v1f + m2v2f, can be used to solve this problem.
To start, let's assign positive directions: west as negative and east as positive. Since car #1 is moving west, its initial velocity (v1i) is -2.5 m/s, and car #2's initial velocity (v2i) is +3.7 m/s.
Now we can substitute the given values into the equation:
(254 kg)(-2.5 m/s) + (254 kg)(3.7 m/s) = (254 kg)(v1f) + (254 kg)(v2f)
Simplifying the equation, we have:
-635 kg·m/s + 939.8 kg·m/s = 254 kg·v1f + 254 kg·v2f
Now we can solve for the final velocities (v1f and v2f). Since the collision is elastic, we can assume that the two cars exchange their velocities.
-635 kg·m/s + 939.8 kg·m/s = 254 kg·v2f + 254 kg·v1f
Substituting the mass values:
304.8 kg·m/s = 254 kg·v2f + 254 kg·v1f
Since the velocities of the cars exchange, we can write:
304.8 kg·m/s = 254 kg·v1f + 254 kg·v2f
Dividing the equation by 254 kg, we get:
1.2 m/s = v1f + v2f
Now, we also know that the relative velocity between the two cars must be conserved. In this case, the relative velocity is the difference between the velocities of the two cars before and after the collision.
Relative velocity before collision = v2i - v1i
= 3.7 m/s - (-2.5 m/s)
= 6.2 m/s
Relative velocity after collision = v1f - v2f
Substituting the relative velocity values, we have:
6.2 m/s = v1f - v2f
Now we have two equations:
1.2 m/s = v1f + v2f
6.2 m/s = v1f - v2f
We can solve these simultaneous equations using a method such as substitution or elimination. Let's use the elimination method:
Adding the two equations, we eliminate v2f:
1.2 m/s + 6.2 m/s = v1f + v2f + v1f - v2f
7.4 m/s = 2v1f
v1f = 7.4 m/s / 2
v1f = 3.7 m/s
Substituting the value of v1f into one of the equations, we can solve for v2f:
1.2 m/s = v1f + v2f
1.2 m/s = 3.7 m/s + v2f
v2f = 1.2 m/s - 3.7 m/s
v2f = -2.5 m/s
Therefore, the final velocities after the collision are:
Car #1 (v1f) = 3.7 m/s
Car #2 (v2f) = -2.5 m/s
Please keep in mind that we assumed an elastic collision, meaning that both the momentum and kinetic energy are conserved.
You're correct! This problem can be solved using the conservation of momentum. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
The formula you mentioned, m1v1 + m2v2 = m1v1' + m2v2', applies to this situation.
Let's label car #1 as m1 and car #2 as m2. Car #1 is moving to the west with a velocity of 2.5 m/s (v1 = -2.5 m/s) and car #2 is moving to the east with a velocity of 3.7 m/s (v2 = 3.7 m/s).
m1 = m2 = 254 kg
To solve for the final velocities (v1' and v2'), we plug in the given values into the formula:
m1v1 + m2v2 = m1v1' + m2v2'
(254 kg)(-2.5 m/s) + (254 kg)(3.7 m/s) = (254 kg)v1' + (254 kg)v2'
Simplifying the equation:
-635 kg*m/s + 939.8 kg*m/s = 254 kg v1' + 254 kg v2'
304.8 kg*m/s = 254 kg v1' + 254 kg v2'
Divide both sides by 254 kg:
(304.8 kg*m/s) / (254 kg) = v1' + v2'
1.2 m/s = v1' + v2'
Since the two cars collide head-on, they will have the same final speed but opposite in direction. Let's denote this speed as vf.
Therefore,
1.2 m/s = vf + (-vf)
1.2 m/s = 0 m/s
So the final velocities after the collision for car #1 and car #2 are both 0 m/s.