Define (a) momentum, (b) kinetic energy - Leaving Cert Physics - Question 6 - 2010

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

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

where:

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

The principle of conservation of momentum states that the total momentum before an event is equal to the total momentum after the event, provided no external forces act on the system. Mathematically, it can be expressed as:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

where:

• $m_1$ and $m_2$ are the masses,
• $v_{1i}$ and $v_{2i}$ are the initial velocities,
• $v_{1f}$ and $v_{2f}$ are the final velocities of the two objects.

In launching a spacecraft, the principle of conservation of momentum applies during the rocket's launch. As fuel is burned and ejected downwards (backward), the rocket responds by moving upwards (forward). The momentum gained by the rocket is equal to the momentum lost by the expelled fuel, preserving total momentum in the system.

This principle ensures that the launch occurs efficiently by maximizing thrust while minimizing waste.

For the first skater (mass = 50 kg, speed = 6 m/s):

$p_1 = m_1 v_1 = 50 \text{ kg} \times 6 \text{ m/s} = 300 \text{ kg m/s}$

For the second skater (mass = 70 kg, speed = 0 m/s):

$p_2 = m_2 v_2 = 70 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg m/s}$

Thus, the momentum of each skater before the collision is:

• Skater 1: 300 kg m/s
• Skater 2: 0 kg m/s.

The momentum of the combined skaters after the collision can be calculated using conservation of momentum.

Since the total momentum before the collision is:

$p_{total ext{, before}} = p_1 + p_2 = 300 + 0 = 300 \text{ kg m/s}$

Thus, the momentum after the collision is also:

$p_{total ext{, after}} = 300 \text{ kg m/s}$

The total mass after the collision is:

$m_{total} = m_1 + m_2 = 50 \text{ kg} + 70 \text{ kg} = 120 \text{ kg}$

Using the momentum calculated, we find the speed:

$300 = m_{total} v_{after}$

Substituting the values:

$300 = 120 v_{after}$

Solving for $v_{after}$:

$v_{after} = \frac{300}{120} = 2.5 \text{ m/s}$

For skater 1:

$KE_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 50 \times (6)^2 = 900 \text{ J}$

For skater 2:

$KE_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \times 70 \times (0)^2 = 0 \text{ J}$

Using the speed found after the collision:

$KE_{total ext{, after}} = \frac{1}{2} m_{total} v_{after}^2 = \frac{1}{2} \times 120 \times (2.5)^2 = 375 \text{ J}$

Before the collision, the total kinetic energy was:

$KE_{total ext{, before}} = KE_1 + KE_2 = 900 + 0 = 900 \text{ J}$

After the collision, the total kinetic energy is:

$KE_{total ext{, after}} = 375 \text{ J}$

This shows that kinetic energy is not conserved during the collision, as some of it was converted to other forms of energy (like heat and sound) during the impact.