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3 years ago

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A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc

ity of the carts after the collision? What type of collision is this?

1 answer:

vredina [299]3 years ago

3 0

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

1)

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 6,000 kg$ is the mass of the first train

$u_1 = 10 m/s$ is the initial velocity of the first train

$m_2 = 4,000 kg$ is the mass of the second train

$u_2 = 0$ is the initial velocity of the second train (initially at rest)

$v$ is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s$

2)

There are two types of collisions:

Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

$K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J$

After:

$K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J$

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

Learn more about momentum here:

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