3. The following apparatus is set up to investigate the law of conservation of linear momentum - Scottish Highers Physics - Question 3 - 2016

To calculate the final velocity ($v$) of the combined vehicles after the collision, we use the conservation of momentum principle:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$

Where:

• $m_1 = 0.85$ kg (mass of vehicle X)
• $m_2 = 0.25$ kg (mass of vehicle Y)
• $u_1 = 0.55$ m/s (initial speed of vehicle X)
• $u_2 = -0.30$ m/s (initial speed of vehicle Y)

Substituting the values: $0.85 imes 0.55 + 0.25 imes (-0.30) = (0.85 + 0.25) v$

Calculating the left side: $0.4675 - 0.075 = (1.1) v$ $0.3925 = 1.1 v$

Now, solving for $v$: $v = \frac{0.3925}{1.1} \approx 0.3564 \text{ m/s}$

To show that the collision is inelastic, we need to compare the total kinetic energy before and after the collision.

Before the collision:

• Kinetic energy of vehicle X: $KE_x = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (0.85)(0.55)^2 = 0.1294 \text{ J}$
• Kinetic energy of vehicle Y: $KE_y = \frac{1}{2} m_2 u_2^2 = \frac{1}{2} (0.25)(0.30)^2 = 0.01125 \text{ J}$
• Total kinetic energy before: $KE_{total, before} = KE_x + KE_y = 0.1294 + 0.01125 = 0.14065 \text{ J}$

After the collision:

• Total mass after collision: $m_{total} = m_1 + m_2 = 0.85 + 0.25 = 1.1 ext{ kg}$
• Kinetic energy after: $KE_{after} = \frac{1}{2} m_{total} v^2 = \frac{1}{2} (1.1)(0.3564)^2 \approx 0.07073 \text{ J}$

Since the total kinetic energy before (0.14065 J) is greater than after (0.07073 J), kinetic energy is lost, thus proving the collision is inelastic.