Thirisha Manikandan

2022-07-19

A car heading north collides at an intersection with a truck of the same mass as the car heading east. If they lock together and travel at 28 m/s at 46° north of east just after the collision, how fast was the car initially traveling? Assume that any other unbalanced forces are negligible.

nick1337

To solve this problem, we can use the principles of conservation of momentum and trigonometry. Let's break it down into steps:
Let the initial velocity of the car be represented by ${v}_{c}$ and the initial velocity of the truck be represented by ${v}_{t}$.
After the collision, the car and truck lock together and travel at a speed of 28 m/s at an angle of 46° north of east. This can be represented by the resulting velocity vector .
To find the initial velocity of the car, we need to determine the magnitude and direction.
1. Magnitude of the resulting velocity:
The magnitude of the resulting velocity ${\stackrel{\to }{v}}_{\text{result}}$ is given as 28 m/s.
2. Direction of the resulting velocity:
The direction of the resulting velocity is 46° north of east. We can represent this as an angle measured counterclockwise from the positive x-axis.
To find the initial velocity of the car, we can use the conservation of momentum. Since no other unbalanced forces are present, the total momentum before the collision is equal to the total momentum after the collision.
The momentum before the collision is given by:
${\stackrel{\to }{p}}_{\text{before}}={m}_{c}{\stackrel{\to }{v}}_{c}+{m}_{t}{\stackrel{\to }{v}}_{t}$
The momentum after the collision is given by:
${\stackrel{\to }{p}}_{\text{after}}=\left({m}_{c}+{m}_{t}\right){\stackrel{\to }{v}}_{\text{result}}$
Equating the two expressions and solving for ${v}_{c}$, we have:
${m}_{c}{\stackrel{\to }{v}}_{c}+{m}_{t}{\stackrel{\to }{v}}_{t}=\left({m}_{c}+{m}_{t}\right){\stackrel{\to }{v}}_{\text{result}}$
To simplify the equation, we can split the velocities into their x and y components:
${m}_{c}{v}_{{c}_{x}}\stackrel{\to }{i}+{m}_{c}{v}_{{c}_{y}}\stackrel{\to }{j}+{m}_{t}{v}_{{t}_{x}}\stackrel{\to }{i}+{m}_{t}{v}_{{t}_{y}}\stackrel{\to }{j}=\left({m}_{c}+{m}_{t}\right){v}_{\text{result}}\mathrm{cos}\left(46°\right)\stackrel{\to }{i}+\left({m}_{c}+{m}_{t}\right){v}_{\text{result}}\mathrm{sin}\left(46°\right)\stackrel{\to }{j}$
By comparing the components, we get two equations:
1. ${m}_{c}{v}_{{c}_{x}}+{m}_{t}{v}_{{t}_{x}}=\left({m}_{c}+{m}_{t}\right){v}_{\text{result}}\mathrm{cos}\left(46°\right)$ (equation 1)
2. ${m}_{c}{v}_{{c}_{y}}+{m}_{t}{v}_{{t}_{y}}=\left({m}_{c}+{m}_{t}\right){v}_{\text{result}}\mathrm{sin}\left(46°\right)$ (equation 2)
We know that the car is heading north and the truck is heading east. Therefore, the y-component of the car's velocity is positive, and the x-component of the truck's velocity is positive. Hence, ${v}_{{c}_{y}}={v}_{c}$ and ${v}_{{t}_{x}}={v}_{t}$.
Substituting these values into equations 1 and 2, we get:
${m}_{c}{v}_{c}+{m}_{t}{v}_{t}=\left({m}_{c}+{m}_{t}\right){v}_{\text{result}}\mathrm{cos}\left(46°\right)$ (equation 3)
${m}_{c}{v}_{c}+{m}_{t}{v}_{t}=\left({m}_{c}+{m}_{t}\right){v}_{\text{result}}\mathrm{sin}\left(46°\right)$ (equation 4)
Solving equations 3 and 4 simultaneously will give us the initial velocity of the car, ${v}_{c}$.

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