Redo Problem P21, this time using the concept of the center-of-momentum reference frame.

A car of mass 2300 kg collides with a truck of mass 4300 kg, and just after the collision the car and truck slide along, stuck together, with no rotation. The car’s velocity just before the collision was⟨38, 0, 0⟩m/s, and the truck’s velocity just before the collision was⟨−16, 0, 27⟩m/s. (a) Your first task is to determine the velocity of the stuck-together car and truck just after the collision. What system and principle should you use? (1) Energy Principle (2) Car plus truck (3) Momentum Principle (4) Car alone (5) Truck alone (b) What is the velocity of the stuck-together car and truck just after the collision? (c) In your analysis in part (b), why can you neglect the effect of the force of the road on the car and truck? (d) What is the increase in internal energy of the car and truck (thermal energy and deformation)? (e) Is this collision elastic or inelastic?

The given data can be listed below,

• The mass of the car is,${\mathrm{m}}_{\mathrm{c}}=2300\mathrm{kg}$ .
• The mass of the truck is, $m_t=4300kg$.
• The velocity of car before collision is, ${\mathrm{u}}_1=38\hat{\mathrm{i}}\mathrm{m}/\mathrm{s}$.

The velocity of the truck is,${\mathrm{u}}_2=-16\hat{\mathrm{i}}+27\hat{\mathrm{k}}\mathrm{m}/\mathrm{s}$

The colliding items must stick together and not separate in a completely inelastic collision. In general, an inelastic collision occurs when two objects collide at a slower rate than when they first collided.

When collision occurred both car and truck will stick together without any friction which happens in inelastic collision. So, the system will only obey law of conservation of momentum and velocity can also be find by using it.

Thus, option 2 and 3 are correct.

The total momentum of the system before collision is given by.

${\stackrel{\rightharpoonup }{p}}_i=m_c{\stackrel{\rightharpoonup }{u}}_1+m_c{\stackrel{\rightharpoonup }{u}}_2$

Here,$m_c$is the mass of the car, $m_t$is the mass of the truck, ${\stackrel{\rightharpoonup }{u}}_1$is the initial velocity of the car and ${\stackrel{\rightharpoonup }{u}}_2$is the initial velocity of the truck.

The total momentum after collision is given by,

${\stackrel{\rightharpoonup }{p}}_f=\left(m_c+m_t\right){\stackrel{\rightharpoonup }{v}}_f$

Here, ${\stackrel{\rightharpoonup }{v}}_f$is the combined velocity of the system after collision.

The velocity of truck-car after collision is given by,

$v_f=\frac{m_c{\stackrel{\rightharpoonup }{u}}_1+m_c{\stackrel{\rightharpoonup }{u}}_2}{\left(m_c+m_t\right)}$

Substitute all the values in the above,

Thus, the velocity of the stuck-together car and truck just after the collision is .

The gravitational force acting on the system will cancel out with normal force of the system by the road as the time for frictional force is short it can be neglected.

Thus, the effect of the friction force of the road on the car and truck can be neglected.

the total kinetic energy of the system before collision is given by,

$K_{before}=\frac{1}{2}{m}_c{u}_1^2+\frac{1}{2}{m}_t{u}_2^2$

Here, $m_c$is the mass of the car, $u_1$is the mass of the truck, is the initial velocity of the car whose magnitude $38m/s$is and $u_2$is the initial velocity of the truck whose magnitude is $31.4\mathrm{m}/\mathrm{s}$.

Substitute all the values in the above,

${\mathrm{K}}_{\mathrm{before}}=\left[\frac{1}{2}\left(2300\mathrm{kg}\right){\left(38\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}\left(4300\mathrm{kg}\right){\left(31.4\mathrm{m}/\mathrm{s}\right)}^2\right]\left(\frac{1\mathrm{J}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right)$

Kbefore=122300kg38m/s2+124300kg31.4m/s21J1kg.m/s2 $=3.76\times {10}^6\mathrm{J}$

The total kinetic energy of the system after collision is given by,

${\mathrm{K}}_{\mathrm{after}}=\frac{1}{2}\left({\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{t}}\right){\mathrm{v}}_{\mathrm{f}}^2$

Here,is the final velocity of the car-truck system whose magnitude is 17.8 m/s.

Substitute values in the above,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{before}}=\left[\frac{1}{2}\left(2300\mathrm{kg}+4300kg\right){\left(17.8m/s\right)}^2\right]\left(\frac{1\mathrm{J}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =1.05\times {10}^6\mathrm{J} \end{gathered}$$

The change in internal energy of the system is given by,

role="math" localid="1657863827031" $∆{\mathrm{E}}_{\mathrm{int}}={\mathrm{K}}_{\mathrm{before}}-{\mathrm{K}}_{\mathrm{after}}$

Here, ${\mathrm{K}}_{\mathrm{before}}$is the kinetic energy before collision, and $K_{after}$is the kinetic energy after collision.

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{int}}=3.76\times {10}^6\mathrm{J}-1.05\times {10}^6\mathrm{J} \\ =2.71\times {10}^6\mathrm{J} \end{gathered}$$

Thus, the increase in the internal energy is$2.71\times {10}^6\mathrm{J}$.

The car and the truck will stick together after collision.

Thus, the collision is inelastic.