A freight car is moving freely along a railroad track at 7.0 m/s and collides with a tanker car that is at rest. After the collision, the two cars stick together and continue to move down the track. What is the magnitude of the final velocity of the cars if the freight car has a mass of 1,200 kg and the tanker car has a mass of 1,600 kg?
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Conservation of momentum applies
momentum before=momentum after.
1,200*7+0=(1,200+1600)V
solve for V
3 m/s
3m/s
To find the magnitude of the final velocity of the cars after the collision, we can use the principle of conservation of linear momentum.
The linear momentum before the collision is equal to the linear momentum after the collision.
Linear momentum (p) is given by the product of mass (m) and velocity (v):
p = mv
Let's denote the final velocity of the cars as V.
Before the collision:
The freight car's initial velocity (v1) = 7.0 m/s
The freight car's mass (m1) = 1,200 kg
The tanker car's initial velocity (v2) = 0 m/s (since it's at rest)
The tanker car's mass (m2) = 1,600 kg
The total linear momentum before the collision is:
p_initial = (m1 * v1) + (m2 * v2)
= (1,200 kg * 7.0 m/s) + (1,600 kg * 0 m/s)
= 8,400 kg*m/s
After the collision, the two cars stick together and move with a common final velocity (V).
The total linear momentum after the collision is:
p_final = (m1 + m2) * V
Since the principle of conservation of linear momentum states that the initial and final linear momenta are equal, we have:
p_initial = p_final
Therefore:
8,400 kg*m/s = (1,200 kg + 1,600 kg) * V
Simplifying the equation:
8,400 kg*m/s = 2,800 kg * V
Dividing both sides of the equation by 2,800 kg:
3 kg*m/s = V
So, the magnitude of the final velocity of the cars after the collision is 3 m/s.