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Chapter 7: Q1MCQ (page 170)

A truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation?

(a) The truck has the greater change of momentum because it has the greater mass.

(b) The car has the greater change of momentum because it has the greater speed.

(c) Neither the car nor the truck changes its momentum in the collision because momentum is conserved.

(d) They both have the same change in magnitude of momentum because momentum is conserved.

(e) None of the above is necessarily true.

Short Answer

Expert verified

(d) They both have the same change in magnitude of momentum because momentum is conserved.

Step by step solution

01

Momentum of the truck and the small car

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.

02

Given data

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

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

03

Explanation for change in momentum

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.

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Most popular questions from this chapter

Billiard balls A and B, of equal mass, move at right angles and meet at the origin of an xy coordinate system as shown in Fig. 7–36. Initially ball A is moving along the y axis at \({\bf{ + 2}}{\bf{.0}}\;{\bf{m/s}}\), and ball B is moving to the right along the x axis with speed \({\bf{ + 3}}{\bf{.7}}\;{\bf{m/s}}\).After the collision (assumed elastic), ball B is moving along the positive y axis (Fig. 7–36) with velocity What is the final direction of ball A, and what are the speeds of the two balls?

FIGURE 7-36

Problem 46. (Ball A after the Collison is not shown)

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