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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck are moving together at 22.3 m/s. How fast was the train going to begin with? Assume there is no friction. My frame of reference sets east a positive​

User J B
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1 Answer

4 votes

Answer:

24.084 m/s

Step-by-step explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity


M_(truck)V_(truck)=V_(common)*(M_(truck) +M_(standing)) where
M_(truck) is the mass of the truck,
V_(truck) is velocity of the truck,
V_(common) is the common velocity of moving and standing truck after collision and
M_(standing) is the mass of the standing truck

Making
V_(truck) the subject we obtain


V_(truck)=\frac { V_(common)*(M_(truck) +M_(standing))}{M_(truck)}

Substituting
M_(truck) as 25000 Kg,
V_(common) as 22.3 m/s,
M_(standing) as 2000 Kg we obtain


V_(truck)=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

User Daniel Guillamot
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