Question

A small pickup truck that has a camper shell slowly coasts toward a red light with negligible friction. Two dogs in the back of the truck are moving and making various inelastic collisions with each other and the walls. What is the effect of the dogs on the motion of the center of mass of the system (truck plus entire load)? What is their effect on the motion of the truck?

Short answer

There will be no effect on the motion of the center of mass of the system and motion of the truck.

Step-by-step solution

Definition of the center of mass

The unique location where the weighted relative position of the distributed mass adds to zero is the centre of mass of a mass distribution in space. A force can be applied to this point to induce a linear acceleration without causing an angular acceleration.

The displacement and velocity of the center of mass of the system when different parts of the system are moving with different velocities or having different displacement is given by

\(\begin{aligned}{x_{com}} &= \dfrac{{\sum\limits_{n = 1}^{n = n} {{m_n}{x_n}} }}{{\sum\limits_{n = 1}^{n = n} {{m_n}} }}\\{v_{com}} &= \dfrac{{\sum\limits_{n = 1}^{n = n} {{m_n}{v_n}} }}{{\sum\limits_{n = 1}^{n = n} {{m_n}} }}\end{aligned}\)

Where\({x_{com}}\) is the displacement of the center of mass and \({v_{com}}\) is the velocity of the center of mass.

Explaining when there is no effect on the motion of the center of mass

In the given situation, there are three parts (two dogs and pickup truck) of the system which are having different velocities and displacements.

Lets assume the mass, displacement and velocity of these three parts

Mass of the first dog\( = {m_1}\)

Mass of the second dog\( = {m_2}\)

Mass of the pickup truck\( = M\)

Velocity of the pickup truck\( = v\)

Displacement of the pickup truck\( = x\)

Velocity of the first dog\( = {v_1}\)

Displacement of the first dog\( = {x_1}\)

Velocity of the second dog\( = {v_2}\)

Displacement of the second dog\( = {x_2}\)

Net displacement and velocity of the center of mass can be written as

\(\begin{aligned}{x_{com}} &= \dfrac{{\sum\limits_{n = 1}^{n = 3} {{m_n}{x_n}} }}{{\sum\limits_{n = 1}^{n = 3} {{m_n}} }} &= \dfrac{{M{x_{truck}} + {m_1}{x_1} + {m_2}{x_2}}}{{M + {m_1} + {m_2}}}\\{v_{com}} &= \dfrac{{\sum\limits_{n = 1}^{n = 3} {{m_n}{v_n}} }}{{\sum\limits_{n = 1}^{n = 3} {{m_n}} }} &= \dfrac{{M{v_{truck}} + {m_1}{v_1} + {m_2}{v_2}}}{{M + {m_1} + {m_2}}}\end{aligned}\)

Normally mass of the pickup truck is very greater than the mass of the dog.So, the mass of the dogs can be ignored in comparison to the mass of the pickup truck.Therefore, applying this assumption you will get

\(\begin{aligned}M > > > {m_1}\& {m_2}\\{x_{com}} &= \dfrac{{M{x_{truck}}}}{M} &= {x_{truck}}\\{v_{com}} &= \dfrac{{M{v_{truck}}}}{M} &= {v_{truck}}\end{aligned}\)

Hence, there will be no effect of the movement of the dogs and their collision on the motion of the center of mass and on the motion of the truck.

Step 3: Motion of the truck  

As mass of the truck is very large as compare to the mass of the dogs. Therefore, there will be no effect of inelastic collision of dogs and their movement on the motion of the pickup truck. The truck moves with the same initial speed.