Crash Momentum: 1500kg Boulder, 500kg Truck

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SUMMARY

The forum discussion centers on a physics problem involving momentum conservation during a collision between a 1500 kg boulder and a 500 kg truck. The boulder moves at 10 m/s westward and collides with the truck traveling at 20 m/s at S30°W. After the collision, the boulder moves at 5 m/s at S70°W. The user attempts to solve for the truck's speed and direction post-collision using horizontal and vertical momentum equations, ultimately concluding that the truck's angle should be 20°.

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Homework Statement


A boulder that weighs 1500 kg is moving 10m/s in a Westerly direction, until it collides with a truck of mass 500kg traveling at 20m/s at a direction S30°W. After the collision the boulder moves away at 5m/s at a direction of S70°W. Find the speed and direction of the truck adter the crash.

Homework Equations


The Attempt at a Solution


I tried sorting the problem into horizontal and vertical aspects.

the horizontal equation looks like:

1500 * -10 + 500(-20 cos 60) = 1500 * (-5 cos 30°) + 500(v cos θ)

which solves to: v cos = -34.38

where the 1500 and the 500 are the masses, the (-10) and (-5 cos 30) are the before the crash horizontal vectors, and (-20 cos 60) and (v cos θ) the after crash horiz vectors.

the vertical equation:

1500 * 0 + 500(-20 sin 60) = 1500 * (-5 sin 30) + 500( v sin θ)

which solves to: v sin θ = -9.82

where 1500 and 500 are the masses again, (0) and (-20 sin 60) the before the crash vertical components, andd (-5 sin 30) & (v sin θ) the after the crash components.

from here i tried:

tan θ = -34.38 / -9.82
θ = 0.061

but I am fairly certain this is wrong. Can anyone please tell me where i went wrong? thank you!
 
Last edited:
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Tankertert said:
[

The Attempt at a Solution


I tried sorting the problem into horizontal and vertical aspects.

the horizontal equation looks like:

1500 * -10 + 500(-20 cos 60) = 1500 * (-5 cos 30°) + 500(v cos θ)

which solves to: v cos = -34.38

!

It should be 20°.
 

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