A car of mass 120 kg, moving to the right at a velocity of 25 m·s⁻¹, collides with the back of a construction vehicle loaded with cement bags and moving in the same direction at a velocity of 6.25 m·s⁻¹ - NSC Technical Sciences - Question 4 - 2021 - Paper 1

The total linear momentum of an isolated system remains constant (i.e., is conserved) in magnitude and direction, provided no external forces act on the system.

Since the question states that the system is isolated, the net external force acting on the system is zero:

$F_{net} = 0 ext{ N}$

Using the principle of conservation of momentum:

$ext{Initial momentum} = ext{Final momentum}$

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$

Substituting the known values:

$(1 ext{ 120} ext{ kg})(25 ext{ m/s}) + (m_{c})(6.25 ext{ m/s}) = (1 ext{ 120} ext{ kg})(7.45 ext{ m/s}) + (m_{c} + 100)(8.45 ext{ m/s})$

Solving the equation gives:

The mass of the construction vehicle, $m_{c}$, is approximately 8345.55 kg.

To determine if the collision is elastic, we check if kinetic energy before and after the collision is conserved:

• Initial kinetic energy: KE_{initial} = rac{1}{2} m_{1} v_{1i}^2 + rac{1}{2} m_{2} v_{2i}^2

• Final kinetic energy: KE_{final} = rac{1}{2} m_{1} v_{1f}^2 + rac{1}{2} m_{2} v_{2f}^2

Calculating these values:

• If $KE_{initial} eq KE_{final}$, then the collision is inelastic.