91影视

Mass of car A is $m_A=2000鈥�kg$.

Mass of car B is$m_B=1500鈥�kg$

If the +x-axis be toward the east and +y-axis toward the north, then the component of velocities are given as-

$$\begin{gathered} v_{A1x}=v_{A1} \\ {v}_{A1y}=0 \\ {v}_{B1x}=0 \\ {v}_{B1x}=-15鈥�m/s \\ {v}_{2x}=v_2\cos 65掳 \\ {v}_{2y}=v_2\sin 65掳 \end{gathered}$$

Here, role="math" localid="1665052882028" $v_{A1x}and{v}_{A1y}$are the x and y component respectively, of velocity of car A before the collision $\left(v_{A1}\right)$; role="math" localid="1665052982472" $v_{B1x}and{v}_{B1y}$ are the x and y component respectively, of velocity of car B before the collision $\left(v_{B1}\right)$and $v_{2x}and{v}_{2y}$are the x and y component respectively, of the common velocity of both the cars $\left(v_2\right)$after the collision.

The law of conservation of momentum states that for an explosion of the stationary body, the sum of the momentum of all the parts of the body, after the explosion, should be zero.

By applying law of conservation of momentum.

$m_A{v}_{A1}+m_B{v}_{B1}=(m_A+m_B)v_2鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�......\left(1\right)$

For two-dimensional motion, we apply the law of conservation of momentum along the x and y-axis separately.

role="math" localid="1665053242157" $$\begin{gathered} m_A{v}_{A1x}+m_B{v}_{B1x}=(m_A+m_B)v_{2x}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�......\left(2\right) \\ {m}_A{v}_{A1y}+m_B{v}_{B1y}=(m_A+m_B)v_{2y}鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�......\left(3\right) \end{gathered}$$

Substituting the given values in equation (2)

$$\begin{gathered} \left((2000鈥�kg)(0鈥�m/s)+(1500鈥�kg)(-15鈥�m/s)=(2000鈥�kg)(-v_2\sin 65) \\ +(1500鈥�kg)(-v_2\sin 65)\right) \\ {v}_2=\frac{1500鈥�kg脳(-15鈥�m/s)}{-\sin 65(2000鈥�kg+1500鈥�kg)} \\ {v}_2=7.09鈥�m/s \end{gathered}$$

Hence, the speed of enmeshed car moving just after collision is 7.09 m/s.

Substitute the values in equation (3) gives;

$$\begin{gathered} \left((2000鈥�kg)v_{A1}+(1500鈥�kg)\left(0\right)=(2000鈥�kg)\left(\right(7.09鈥�m/s)\cos 65掳) \\ +(1500鈥�kg)\left(\right(7.09鈥�m/s)\cos 65掳)\right) \\ {v}_{A1}=\frac{(2000鈥�kg)(7.09\cos 65掳)+(1500鈥�kg)(7.09\cos 65掳)}{2000kg} \\ {v}_{A1}=5.25鈥�m/s \end{gathered}$$

Hence, the speed of car A going just before collision is 5.25 m/s .