Question

Refer to the following scenario for the next 2 questions: A car (mass 1000 kg) moving at 10 m/s east collides head on with a train (mass 20,000 kg) moving at 5 m/s west. The car and train stick together after the collision.

What is the final velocity of the car-train system? (Define east to be the positive direction.) *

4.3 m/s

-4.3 m/s

5.2 m/s

-5.2 m/s

What is the change in kinetic energy in this collision? *

The system gains 16000 J of KE

The system loses 16000 J of KE

The system gains 106000 J of KE

The system loses 106000 J of KE

KE is conserved, so it does not change

Answer 1

Let $m_T, m_C$ and $v_T, v_C$ be the masses and initial velocities of train and car respectively. Let's assume that the car-train system will move west after the collision with velocity $v$.

Hence, according to the conservation of momentum:

$m_C v_C - m_T v_T = - (m_C + m_T) v$,

from where $v = \frac{m_T v_T - m_C v_C}{m_C + m_T} \approx 4.28 \frac{m}{s}$.

Since the east is the positive direction, final velocity is approximately $-4.3 \frac{m}{s}$.

The collision is inelastic, hence kinetic energy will not be conserved. The change is kinetic energy is:

$\Delta E = \frac{(m_C + m_T)v^2}{2} - \left( \frac{m_C v_C^2}{2}+ \frac{m_T v_T^2}{2} \right) \approx -106000 J$ - system loses $106000 J$ of its kinetic energy.