- #1
erisedk
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Homework Statement
A pickup truck is driving at night through the rain. He drives onto a frozen lake. He immediately turns off his engine to save fuel (it wouldn't help anyway on frictionless ice) and let's the truck coast (move under no power). Unfortunately, the rain is accumulating in the back of his truck, at a constant rate of σ kg/sec. The given parameters of this problem are the truck's initial mass m0 and velocity v0 xˆ when it hits the lake at time t = 0, and the rate σ of rain accumulation. Also, the rain is falling straight downwards in the reference frame of the ground.
(a) Calculate the truck's speed v(t) as a function of time.
(b) New situation: As soon as he hits the frozen lake at t = 0, the trucker realizes that he can help himself by bailing the rain from the back of his truck. The trucker is very fit: he is able to get rid of every raindrop as it
falls by catching it in a bucket and hurling it backwards off the truck (–x direction) at speed u (relative to the truck). Calculate the truck's speed v(t) in this new situation.
(c) Final situation: This time, the trucker has some good luck → the rain is falling mostly downward but it also has a horizontal speed component of u (relative to the ground) in the +x direction. The trucker let's the rain accumulate as in part (a), hoping that the rain's forward speed component will help push him across the lake. Calculate the truck's speed v(t) in this last situation.
Homework Equations
Conservation of momentum
The Attempt at a Solution
I just feel like my answers are wrong, because they're not like the rocket problems that we get, and I didn't use any differentials or calculus of any sort.
(a) As there are no external forces in the x-direction, the momentum will be conserved in the x-direction.
Here, the system includes the truck and the rain.
##m_0v_0 = (m_0 + \sigma t)v##
## v = \dfrac{m_0v_0}{(m_0 + \sigma t)}##
(b) As there are no external forces in the x-direction, the momentum will be conserved in the x-direction.
##m_0v_0 = m_0v + (\sigma t)(v-u)##
##v = \dfrac{m_0v_0 + \sigma tu}{m_0 + \sigma t}##
(c) Again, conserving momentum in the x direction-
##m_0v_0 + \sigma tu = (m_0 + \sigma t)v##
##v = \dfrac{m_0v_0 + \sigma tu}{m_0 + \sigma t}##
Please help!