A 12000 kg train engine moving at 2.2 m/s locks into 3 boxcars with a total mass of 25000 kg sitting still. If the collision is inelastic and has little friction, how fast will the engine and cars move?
I got 1.4 but correct answer is 0.71?
I don't understand how
12000x2.2 +25000x0 =26400
12000+25000=37000/26400
=1.4
momentum is conserved
(12000 * 2.2) + (25000 * 0) = (12000 + 25000) * v
so I should set this up at initialV=combined and divided by the combined mass
That's where I'm messing it up.
thanks
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of a closed system before a collision is equal to the total momentum after the collision.
Before the collision:
- The train engine has a mass of 12000 kg and is moving at a speed of 2.2 m/s. Therefore, its momentum is given by: momentum_engine = mass_engine x velocity_engine = 12000 kg x 2.2 m/s = 26400 kg·m/s.
- The boxcars have a total mass of 25000 kg and are stationary, so their initial momentum is zero: momentum_boxcars = mass_boxcars x velocity_boxcars = 25000 kg x 0 m/s = 0 kg·m/s.
After the collision, the train engine and boxcars stick together and move as a single unit. Let's call their combined velocity after the collision V.
Using the principle of conservation of momentum, we can set up the equation:
total_initial_momentum = total_final_momentum
momentum_engine + momentum_boxcars = (mass_engine + mass_boxcars) x V
Substituting the given values:
26400 kg·m/s + 0 kg·m/s = (12000 kg + 25000 kg) x V
26400 kg·m/s = 37000 kg x V
Solving for V:
V = 26400 kg·m/s / 37000 kg = 0.71 m/s (rounded to two decimal places)
Therefore, the final velocity of the engine and boxcars, when they move as a single unit after the collision, is approximately 0.71 m/s.