Arthur Pratt

2021-12-19

A car with mass is traveling west through an intersection at a magnitude of velocity of vc = 15 m/s when a truck of mass traveling south at fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of ${\mu }_{k}=0.5$.
a)Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and j.
b)How far, in meters, will the vehicles slide after the collision?

Mary Herrera

Given information:
The mass of the car
The velocity of the car (consider east as +ve x direction or +ve i)
The mass of the truck
The velocity of the truck
According to law of conservation of momentum the momentum in all the direction is conserved, considering the along the "i" direction we can write:
${m}_{c}{v}_{c}+{m}_{t}{v}_{t}=\left({m}_{c}+{m}_{t}\right){V}_{i}\to {V}_{i}=-\left(\frac{{m}_{c}}{{m}_{c}+{m}_{t}}\right){v}_{c}$
Now, consider the momentum conservation along the "j" direction:
${m}_{c}{v}_{c}+{m}_{t}{v}_{t}=\left({m}_{c}+{m}_{t}\right){V}_{j}\to {V}_{j}=-\left(\frac{{m}_{t}}{{m}_{c}+{m}_{t}}\right){v}_{t}$
Therefore, the velocity of the vehicles after the collsion is:
$V=-\left(\frac{{m}_{c}}{{m}_{c}+{m}_{t}}\right){v}_{c}\stackrel{^}{i}-\left(\frac{{m}_{t}}{{m}_{c}+{m}_{t}}\right){v}_{t}\stackrel{^}{j}$
By substituting the correspondingvalues, we get the velocity of the vehciles after the collision as:

After the collision the cinetic energy of the both vehicles combined will be utilized to do work against the dricion force (given ${\mu }_{k}=0.5$), The friction force both the vehicles is given by:

Now, according to work energy theorem the we can write:
$\frac{1}{2}\left({m}_{c}+{m}_{t}\right){V}^{2}=f.d$
Where "d" is the distance travelled by the wreckage after the collision. Substituting the corresponding values, we get:

Hence, the distance travelled by the wreckage after the collsion is 7.958 m.

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