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daser333 [38]
3 years ago
12

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​
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

24.084 m/s

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

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igomit [66]

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Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

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