Define (i) momentum and (ii) kinetic energy - Leaving Cert Physics - Question 6 - 2018

Kinetic energy is the energy that an object possesses due to its motion. The formula for kinetic energy is given by:

$KE = \frac{1}{2}mv^2$

where:

• $KE$ is the kinetic energy,
• $m$ is the mass, and
• $v$ is the velocity.

The momentum of car A is calculated as:

$p_A = m_A \cdot v_A = (500 \text{ kg}) \cdot (6 \text{ m s}^{-1}) = 3000 \text{ kg m s}^{-1}$

The momentum of car B is stationary, hence:

$p_B = m_B \cdot v_B = (300 \text{ kg}) \cdot (0 \text{ m s}^{-1}) = 0 \text{ kg m s}^{-1}$

According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision. Therefore, the momentum of the combined cars after the collision is:

$p_{AB} = p_A + p_B = 3000 \text{ kg m s}^{-1} + 0 = 3000 \text{ kg m s}^{-1}$

Using the conservation of momentum:

$p_{AB} = (m_A + m_B) \cdot v_f$

Plugging in the values:

For car A:

$KE_A = \frac{1}{2} m_A v_A^2 = \frac{1}{2} \cdot (500) \cdot (6^2) = \frac{1}{2} \cdot 500 \cdot 36 = 9000 \text{ J}$

For car B (stationary):

$KE_B = \frac{1}{2} m_B v_B^2 = \frac{1}{2} \cdot (300) \cdot (0^2) = 0 ext{ J}$

After the collision, the combined mass (800 kg) moves at a speed of 3.75 m/s. The kinetic energy is:

$KE_{AB} = \frac{1}{2} (m_A + m_B) v_f^2 = \frac{1}{2} \cdot (800) \cdot (3.75^2) = 5625 \text{ J}$

The initial total kinetic energy of both cars is 9000 J + 0 J = 9000 J. After the collision, the total kinetic energy is 5625 J. This shows that kinetic energy is not conserved in inelastic collisions like this one, as some energy is transformed into other forms, such as heat or sound.