91Ó°ÊÓ

In this question, a major misconception that can arise is that the truck's momentum is higher than that of the car because it has more mass.

Momentum is defined as the product of mass and velocity, so it does not only depend on the mass of the truck.

Another misconception that can arise is that a small car has a greater variation in momentum because the change in its speed is higher than that for the truck.

The speed of the truck is\(v = 15\;{\rm{km/h}}\).

The speed of the small car is \(u = 30\;{\rm{km/h}}\).

After a head-on collision, the two bodies move together at the same speed. In this case, zero external forces act on the car or truck, so the system’s momentum is conserved.

By the principle of conservation of momentum, the momentum lost by the truck is equal to that gained by the car.

Assume that \({m_{\rm{T}}}\) is the mass of the truck, \({m_{\rm{C}}}\) is the mass of the car, \({v_{\rm{C}}}\) and \({v_{\rm{T}}}\) are initial velocities of the car and the truck, and \({v'_{\rm{C}}}\) and \({v'_{\rm{T}}}\) are the final velocities of the car and the truck.

Then, using the principle of conservation of momentum, you can write

\(\begin{aligned}{l}{m_{\rm{T}}}{v_{\rm{T}}} + {m_{\rm{C}}}{v_{\rm{C}}} = {m_{\rm{T}}}{{v'}_{\rm{T}}} + {m_{\rm{C}}}{{v'}_{\rm{C}}}\\{m_{\rm{T}}}\left( {{v_{\rm{T}}} - {{v'}_{\rm{T}}}} \right) = {m_{\rm{C}}}\left( {{{v'}_{\rm{C}}} - {v_{\rm{C}}}} \right)\\\left| {{m_{\rm{T}}}\left( {{v_{\rm{T}}} - {{v'}_{\rm{T}}}} \right)} \right| = \left| {{m_{\rm{C}}}\left( {{v_{\rm{C}}} - {{v'}_{\rm{C}}}} \right)} \right|.\end{aligned}\)

It means the magnitude of change in the momentum of the truck is equal to the magnitude of change in the momentum of the car.

Thus, option (d) is the correct answer.