91影视

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, ${渭}_k=0.13$

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:

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

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,

$W=鈭�K$

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

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

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

Since, car A is initially at rest K,

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

Equating equation (1) and (2),

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

Rearranging the equation for,

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

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

To calculate final velocity of car B,

Using equation (3),

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

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 $\stackrel{鈫�}{P_i}$before collision = Total momentum after collision

For the given situation,

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

$\stackrel{鈫�}{P_i}=\stackrel{鈫�}{P_{ia}}+\stackrel{鈫�}{P_{ib}}$

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

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

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

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

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,

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

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

It is assumed that momentum is conserved,

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

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

Hence,$鈭�P鈮�0$, 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.