What will be the velocity of the truck after 50 minutes? This question came in the Rajshahi University admission exam 2019-20
Q) A truck of 5 metric ton filled with sand was moving with a velocity of $20ms^{-1}$. A hole was created in the truck through which $20kg$ sand falls down each minute. If momentum is conserved, then what will be the velocity of the truck after $50$ minutes?
(a) $20ms^{-1}$
(b) $25ms^{-1}$
(c) $25.5ms^{-1}$
(d) None
My attempt:
I think the answer will be (a). I am basing my answer on the conservation of linear momentum. The sand that was ejected still had the same velocity as the truck at the time of the ejection, so even as the sand is being ejected, the truck should still have the same velocity after 50 minutes.
I'll try to show this mathematically.
mass of sand ejected after 50 min, $m_1=20\times50=1000kg$
(mass of remaining sand + mass of truck)$=m_2=5000-1000=4000kg$
velocity of truck before ejection of any sand, $v=20ms^{-1}$
velocity of ejected sand, $v_1=20ms^{-1}$
velocity of truck after ejection of sand, $v_2=?$
According to the law of conservation of momentum,
$$v(m_1+m_2)=m_1v_1+m_2v_2$$
$$v_2=\frac{v(m_1+m_2)-m_1v_1}{m_2}$$
$$v_2=\frac{20(1000+4000)-1000\times 20}{4000}$$
$$v_2=20ms^{-1}\text{(Ans.)}$$
This is why I think that (a) is the answer.
A third party's claim:
A third-party question bank is of the opinion that (b) is the correct answer. The following is their attempt:
initial mass$=5000kg$
After 20 minutes $=(5000-20\times 50)kg=4000kg$
Since momentum is conserved, $5000\times 20=4000\times x \therefore x=25ms^{-1}$. So, (b) is the answer

Isn't the third party wrong?
 A: Your assessment is correct - if the truck moves over a frictionless surface in vacuum:
The sand which falls through the hole does not transfer any momentum to the truck in doing so. Thus the momentum of each remaining part and cargo of the truck remains the same. Mind: this does mean that the total momentum of the truck with its remaining cargo drops as the dropping sand carries away its own momentum as it leaves the truck with the truck's speed and of course its own weight. What does not change here is the specific momentum, that is the momentum per unit mass.
The situation would be different, if someone threw the sand from the back of the truck with a certain velocity. In essence we then would have a rocket. but that's not answer (b) or (c) either and would need an assumption on the velocity used with which it is being thrown out.
Answer (d) would be correct for a real-life truck on a real surface with friction in air without any engine running: the drag of the air and the friction with the ground would dissipate any initial kinetic energy and it would come to a stop. With some guesses on the truck's cross-section and aerodynamic properties as well as the friction of the tires one could calculate the stopping time - but experience tells us that it definitely is less than 50 minutes.
As such: you can argue (a) or (d) - you just have to argue correctly. Or maybe the question or exam in general gives additional information which you did not quot here and allows to argue only (a) or (d). :)
