Question

Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55.9 mph). The cars lock together in such a way as to prevent them from separating or rotating significantly. One second before the collision, the car is located at position (x,y) = (-20,0) m and the truck is located at position (x,y) = (0,-25) m.
1. Calculate the x-coordinate of the center of mass of the two-automobile system, xCM​​, one second before the collision. ________ m
2. Calculate the y-coordinate of the center of mass of the two-automobile system, xCM​​, one second before the collision.________ m
3. Calculate the magnitude of the velocity of the center of mass of the two-automobile system, vCM​​, one second before the collision. ________ m/s
4. Calculate the direction of the velocity of the center of mass of the two-automobile system, vCM​​ one second before the collision. (Give you answer in degrees relative to the positive x-axis which points to the east.)
________ degrees
5. Calculate the magnitude of the velocity of the center of mass of the two-automobile system, vCM, immediately after the collision. ________ m/s
6. Calculate the direction of the velocity of the center of mass of the two-automobile system, vCM immediately after the collision. (Give you answer in degrees relative to the positive x-axis which points to the east.)
________ degrees

Answer 1

Answer:

1. $X_{cm}=-8.57m$

2. $Y_{cm}=-14.29m$

3. $V_{cm} = 16.66m/s$

4. α = 59.05°

5. $V_{cm} = 16.66m/s$

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

$X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}$

where

$X_c=-20m$;  $m_c=1500kg$;  

$X_t=0m$;   $m_t=2000kg$;

Replacing these values:

$X_{cm}=-8.57m$

$Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}$

where

$Y_c=0m$;  $m_c=1500kg$;  

$Y_t=-25m$;   $m_t=2000kg$;

Replacing these values:

$Y_{cm}=-14.29m$

The velocity of their center of mass is:

$V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}$

where

$V_{c-x}=20m/s$;  $m_c=1500kg$;  

$V_{t-x}=0m/s$;   $m_t=2000kg$;

Replacing these values:

$V_{cm-x}=8.57m/s$

$V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}$

where

$V_{c-y}=0m$;  $m_c=1500kg$;  

$V_{t-y}=-25m$;   $m_t=2000kg$;

Replacing these values:

$V_{cm-y}=-14.29m$

So, the magnitude of the velocity is:

$V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}$

$V_{cm}=16.66m/s$

The angle of the velocity is:

$\alpha =atan(V_{cm-y}/V_{cm-x})$

$\alpha=59.05\°$

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east