In the “before” part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg ). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ${\mathbf{\mu }}_{\mathbf{k}}\mathbf{}\mathbf{of}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{13}$ ) stops them, at distances ${\mathbf{d}}_{\mathbf{A}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{m}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{d}}_{\mathbf{B}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{m}$. What are the speeds of (a) car A and (b) car Bat the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision. (d) Explain why this assumption may be invalid.

Mass of the car A,${\mathrm{m}}_{\mathrm{a}}=1100\mathrm{kg}$

Mass of the car B,${\mathrm{m}}_{\mathrm{b}}=1400\mathrm{kg}$

Coefficient of friction, ${\mu }_k=0.13$

Distance covered by car A, $d_a$= 8.2 m

Distance covered by car B,$d_b$ = 6.1 m

By applyingtheprinciple of conservation of momentum and usingtheconcept that work done by the system is change in K.E. of the system, findtherequired values of velocities. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$$\begin{gathered} P_i=P_f \\ P=mv \\ K=\frac{1}{2}m{v}^2 \\ W=∆K \end{gathered}$$

where, m is mass, v is velocity, P is linear momentum, Wis work done and K is kinetic energy.

To find out final speed of car A, change in K.E. of the system is equal to work done by the system,

$W=∆K$

W = work done by frictional force is given by$W=f_k{d}_a$

Frictional force is given by $f_k=\mu m_{a}g$

$$\begin{gathered} W=\mu m_{a}g{d}_a.........\left(1\right) \\ ∆K=K_f-K_i \end{gathered}$$

Since, car A is initially at rest K,

$\begin{array}{l}∆K=K_f\\ ∆K=\frac{1}{2}{m}_a{V}_{fa}^2\dots \dots \left(2\right)\end{array}$

Equating equation (1) and (2),

$\frac{1}{2}{m}_a{V}_{fa}^2=\mu m_{a}g{d}_a$

Rearranging the equation for,

$$\begin{gathered} v_{af}=\sqrt{2\mu g{d}_a}.........\left(3\right) \\ {v}_{af}=\sqrt{\left(2\times 0.13\times 9.8\times 8.2\right)} \\ {v}_{af}=4.57\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the final speed of car A is 4.6 m/s.

To calculate final velocity of car B,

Using equation (3),

$$\begin{gathered} v_{bf}=\sqrt{\left(2\mu g{d}_b\right)} \\ {v}_{bf}=\sqrt{\left(2\times 0.13\times 9.8\times 6.1\right)} \\ {v}_{bf}=3.94\mathrm{m}/\mathrm{s} \\ \approx 3.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the final speed of car B is 3.9 m/s.

As, final velocity of car B, ${\mathrm{v}}_{\mathrm{bf}}=3.94\mathrm{m}/\mathrm{s}$, find its initial velocity using principle of conservation of momentum.

Total momentum $\overrightarrow{P_i}$before collision = Total momentum after collision

For the given situation,

Total initial momentum = Initial momentum of car A+ Initial momentum of car B

$\overrightarrow{P_i}=\overrightarrow{P_{ia}}+\overrightarrow{P_{ib}}$

As initially car A is at rest, $\overrightarrow{P_{ia}}=0$

$\overrightarrow{P_i}=\overrightarrow{P_{ib}}=m_b{v}_{ib}..........\left(4\right)$

Total final momentum = final momentum of car A + final momentum of Car B

$\overrightarrow{P_f}=m_a{v}_{fa}+m_b{v}_{bf}........\left(5\right)$

Equating equation (4) and (5),

$m_b{v}_{ib}=m_a{v}_{fa}+m_b{v}_{bf}$

Final velocity of the block can be calculated by,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{ib}}=\frac{{\mathrm{m}}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{fa}}+{\mathrm{m}}_{\mathrm{b}}{\mathrm{v}}_{\mathrm{bf}}}{{\mathrm{m}}_{\mathrm{b}}} \\ {\mathrm{v}}_{\mathrm{ib}}=\frac{1100\times 4.57+1400\times 3.94}{1400} \\ {\mathrm{v}}_{\mathrm{ib}}=\frac{{\mathrm{m}}_{\mathrm{a}}{\mathrm{v}}_{\mathrm{fa}}+{\mathrm{m}}_{\mathrm{b}}{\mathrm{v}}_{\mathrm{bf}}}{{\mathrm{m}}_{\mathrm{b}}} \\ {\mathrm{v}}_{\mathrm{ib}}=7.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the initial velocity of car B is 7.5 m/s.

It is assumed that momentum is conserved,

i.e. $$\begin{gathered} ∆P=0 \\ ∆P=Fdt=0 \end{gathered}$$

Here, force acting is the frictional force $f_k$which is not zero.

Hence,$∆P\ne 0$, and assumption ofconservation of linear momentum may be invalid.

Therefore, by applying the principle of conservation of momentum and using the concept that work done by the system is change in kinetic energy of the system, required values can be found.