91Ó°ÊÓ

Example 8.10

The initial velocity of bock A, $V_{A1}=2.0m/s$

The final velocity of block A, $V_{A2}=-1.0m/s$

The initial velocity of bock B, $V_{B1}=-2.0m/s$

The final velocity of block B, $V_{B2}=3.0m/s$

Example 8.11

The initial velocity of the neutron, $v_{n1}=2.6×{10}^{7}m/s$

The final velocity of the neutron, $v_{n2}=-2.2×{10}^{7}m/s$

The initial velocity of carbon’s nucleus, $v_{c1}=0$

The final velocity of carbon’s nucleus, $v_{c2}=0.4×{10}^{7}m/s$

Example 8.12

The initial velocity of puck A, $\stackrel{→}{v_{A1}}=\left(4.0m/s\right)i$

Thefinal velocity of puck A,

$$\begin{gathered} \stackrel{→}{v_{A2}}=\left(2.00\cos 36.9\stackrel{Áåž}{i}+2.00\sin 36.9\stackrel{Áåž}{j}+1.60\stackrel{Áåž}{1}+1.20\hat{j}\right)m/s \\ =1.60\hat{i}+1.20\hat{j}m/s \end{gathered}$$

The initial velocity of puck B,$\stackrel{→}{v_{B1}}=0$

The final velocity of puck B,

$$\begin{gathered} \stackrel{→}{v_{B2}}=\left(4.47\cos 26.6\hat{i}-4.47\sin 26.6\hat{j}+4.00-2.00j\right)m/s \\ =4.00\hat{i}-2.00\hat{j}m/s \end{gathered}$$

The pace at which an object's position changes is represented by a velocity vector. A velocity vector's magnitude indicates an object's speed, whereas the vector's direction indicates its direction.

We know,

The initial velocity of bock A,$V_{A1}=2.0m/s$

The final velocity of block A,$V_{A2}=-1.0m/s$

The initial velocity of bock B, $V_{B1}=-2.0m/s$

The final velocity of block B, $V_{B2}=3.0m/s$

Relative velocity before the collision

$$\begin{gathered} \stackrel{→}{v_{A1}}-\stackrel{→}{v_{B1}}=2.0-\left(-2.0\right) \\ =4.0m/s \end{gathered}$$

Relative velocity after the collision

$$\begin{gathered} \stackrel{→}{v_{A2}}-\stackrel{→}{v_{B2}}=-1.0m/s-3.0m/s \\ =-4.0m/s \end{gathered}$$

The relative velocity before and after the collision are equal in magnitude but opposite in direction.

We know,

The initial velocity of the neutron,$v_{n1}=2.6×{10}^{7}m/s$

The final velocity of the neutron,$v_{n2}=-2.2×{10}^{7}m/s$

The initial velocity of carbon’s nucleus,$v_{c1}=0$

The final velocity of carbon’s nucleus,$v_{c2}=0.4×{10}^{7}m/s$

Relative velocity before the collision

$$\begin{gathered} \stackrel{→}{v_{n1}}-\stackrel{→}{v_{c1}}=2.6×{10}^{7}m/s-0m/s \\ =2.6×{10}^{7}m/s \end{gathered}$$

Relative velocity after the collision

$$\begin{gathered} \stackrel{→}{v_{n2}}-\stackrel{→}{v_{c2}}=-2.2×{10}^{7}m/s-0.4×{10}^{7}m/s \\ =-2.6×{10}^{7}m/s \end{gathered}$$

The relative velocity before and after the collision are equal in magnitude but opposite in direction.

We know,

The initial velocity of puck A,$\stackrel{→}{v_{A1}}=\left(4.0m/s\right)i$

The final velocity of puck A,

$\stackrel{→}{v_{A2}}=\left(2.00\cos 36.9\hat{i}+2.0\sin 36.9\hat{j}+1.60\hat{i}+1.20\hat{j}\right)m/s$

The initial velocity of puck B,$\stackrel{→}{v_{B1}}=0m/s$

The final velocity of puck B,

$\stackrel{→}{v_{B2}}=\left(4.47\cos 26.6\hat{i}-4.47\sin \hat{j}+26.6\hat{j}+4.00-2.00j\right)m/s$

Relative velocity before the collision

$$\begin{gathered} \stackrel{→}{v_{A1}}-\stackrel{→}{v_{B1}}=\left(4.00m/s\right)i-0m/s=4.00\hat{i}m/s \\ \mathrm{θ}=0 \end{gathered}$$

Relative velocity after the collision

$$\begin{gathered} \stackrel{→}{v_{A2}}-\stackrel{→}{v_{B2}}=\left(1.60\hat{i}+1.20\hat{j}\right)-\left(4.00\hat{i}-2.00\hat{j}\right)m/s \\ =-2.4\hat{i}+3.2m/s \end{gathered}$$

The magnitude of relative velocity after the collision

$$\begin{gathered} =\sqrt{{\left(-2.40m/s\right)}^2+{\left(3.20m/s\right)}^2} \\ =4.0m/s \end{gathered}$$

The direction of relative velocity after the collision

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\frac{3.20}{-2.40} \\ =180-{\tan }^{-1}\frac{3.20}{-2.40} \\ =127° \end{gathered}$$

The relative velocity before and after the collision is equal in magnitude but direction changes by an angle of$127°$ .