Question

A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in. part a what is the car's speed just after the gravel is loaded?

Answer 1

Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s

Answer 2

The car's speed just after the gravel is loaded is 0.952 m/s.

Given that,

• A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in.
• Here we used the conservation of energy and momentum.

Based on the above information, the calculation is as follows:

$m_1 \times v_1=(m_1+m_2)\times v_2$

Now the car speed should be

$= (20000kg \times 1.00m/s)\div (21000kg)$

=.952m/s

Therefore we can conclude that the car's speed just after the gravel is loaded is 0.952 m/s.

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