Question

In each of Examples 8.10, 8.11, and 8.12 (Section 8.4),verify that the relative velocity vector of the two bodies has thesame magnitude before and after the collision. In each case, whathappens to the directionof the relative velocity vector?

Short answer

In the case of Examples 8.10,the relative velocity before and after the collision is equal in magnitude but opposite in direction.

In the case of Examples 8.11,the relative velocity before and after the collision is equal in magnitude but opposite in direction.

In the case of Examples 8.12,The relative velocity before and after the collision are equal in magnitude but direction changes by an angle of.

Step-by-step solution

Given information in the Examples 8.10,8.11 and 8.12 (section 8.4)

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, $v_{n1}=2.6\times {10}^{7}m/s$

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

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

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

Example 8.12

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

Thefinal velocity of puck A,

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

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

The final velocity of puck B,

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

Significance of relative velocity vector

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.

Finding the magnitude and direction of relative velocity for Example 8.10

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

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

Relative velocity after the collision

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

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

Finding the magnitude and direction of relative velocity for Example 8.11

We know,

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

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

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

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

Relative velocity before the collision

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

Relative velocity after the collision

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

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

Finding the magnitude and direction of relative velocity for Example 8.12

We know,

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

The final velocity of puck A,

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

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

The final velocity of puck B,

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

Relative velocity before the collision

$$\begin{gathered} \overrightarrow{v_{A1}}-\overrightarrow{v_{B1}}=\left(4.00m/s\right)i-0m/s=4.00\hat{i}m/s \\ \theta =0 \end{gathered}$$

Relative velocity after the collision

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

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

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

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