Impulse

Classical Mechanics Level 2

There is a train that weighs 100kg and is 30m long. And the train starts to gain weight by 5kg/s after it starts moving. If you drag the train that was at rest with a constant force of 75N, what is the speed of the train after 5 seconds? (ignore the friction)

3m/s 3.347m/s 2.99m/s 3.147m/s

David Ahn by David Ahn , 19, South Korea

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1 solution

David Ahn
Sep 15, 2018
  • F = d p d t = d ( m v ) d t = d m d t v + d v d t m = 5 v + d v d t ( m 0 + 5 t ) F=\frac { dp }{ dt } =\frac { d(mv) }{ dt } =\frac { dm }{ dt } v+\frac { dv }{ dt } m=5v+\frac { dv }{ dt } ({ m }_{ 0 }+5t)
  • d v d t = F − 5 v m 0 + 5 t \frac { dv }{ dt } =\frac { F-5v }{ { m }_{ 0 }+5t } , d t m 0 + 5 t = d v F − 5 v \frac { dt }{ { m }_{ 0 }+5t } =\frac { dv }{ F-5v }
  • ∫ 0 5 d t m 0 + 5 t = ∫ 0 v d v F − 5 v \int _{ 0 }^{ 5 }{ \frac { dt }{ { m }_{ 0 }+5t } } =\int _{ 0 }^{ v }{ \frac { dv }{ F-5v } }
  • 1 5 l n ( m 0 + 25 m 0 ) = − 1 5 l n ( F − 5 v F ) \frac { 1 }{ 5 } ln(\frac { { m }_{ 0 }+25 }{ { m }_{ 0 } } )=-\frac { 1 }{ 5 } ln(\frac { F-5v }{ F } )
  • m 0 + 25 m 0 = F F − 5 v \frac { { m }_{ 0 }+25 }{ { m }_{ 0 } } =\frac { F }{ F-5v }
  • 5 + 25 5 = 75 75 − 5 v \frac { 5+25 }{ 5 } =\frac { 75 }{ 75-5v }
  • ∴ v = 3 m / s \therefore v=3m/s
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