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Paha777 [63]
3 years ago
6

Two cars of the same mass collide at an intersection. Just before the collision, one car is traveling east at 80.0 km/h and the

other car is traveling south at 60.0 km/h. If the collision is completely inelastic, so the two cars move as one object after the collision, what is the speed of the cars immediately after the collision?
Physics
1 answer:
Elden [556K]3 years ago
5 0

Answer:

v_{f}=70\frac{km}{h}

Explanation:

In a completely inelastic collision the momentum P is conserved, so can be expressed as:

P_{f}-P_{i}=0

As P is defined as: P=m.v

Replacing:

m_{1}.v_{1f}+m_{2}.v_{2f}=m_{1}.v_{1i}+m_{2}.v_{2i} (Eq. 1)

As the problem says that the two cars have the same mass:

m_{1}=m_{2}=m

And in a completely inelastic collision the velocities after the collision are equal, so:

v_{1f}=v_{2f}=v_{f}

So replacing in Eq. 1:

m.v_{f}+m.v_{f}=m.v_{1i}+m.v_{2i}

2m.v_{f}=m.(v_{1i}+v_{2i})

Solving for v_{f}:

v_{f}=\frac{(v_{1i}+v_{2i})}{2}

And replacing the values for the velocity:

v_{f}=\frac{(80.0\frac{km}{h}+60.0\frac{km}{h})}{2}

v_{f}=70\frac{km}{h}

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Answer:

Explanation:

Given

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similarly B travels with 20 mph and in 2 hours

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Velocity of B

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v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}

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