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# A cart of mass 150 kg is pulled horizontally on a frictionless surface with force 10 N. If 100$g{{s}^{-1}}$ sand is dropped in the cart vertically then find the speed of the system when cart has 100 kg sand on it.A) 10$m{{s}^{-1}}$B) 20$m{{s}^{-1}}$C) 40$m{{s}^{-1}}$D) 50$m{{s}^{-1}}$

Last updated date: 13th Jun 2024
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Hint:We need to understand the relation between the rate of change of the mass of the cart-sand system with the velocity attained by the system with a constant application of force. The mass determines the acceleration at every instant in this case.

Complete Step-by-Step Solution:
We are given a cart which is of 150kg in mass which is pulled along a frictionless floor in the horizontal direction with a constant force. The velocity of the cart system is supposed to be increasing at a uniform rate in such a case. But in this case, the cart is added with sand at a constant rate all throughout the journey. This increases the mass of the system and therefore, the force applied is not linearly increasing the velocity.

Let us consider $\eta$ as the mass added to cart per second of the journey and m be the mass at any instant. The force which is applied is used to move the mass already present in the cart at an instant and the additional mass added at that instant.

We know that $\eta$ is the rate of change of mass, so, if v is the velocity of the cart at every instant, then the total force distribution can be given as –
\begin{align} & F=\eta v+m\dfrac{dv}{dt} \\ & \therefore F-\eta v=m\dfrac{dv}{dt} \\ \end{align}

Now, we can apply the necessary limits and integrate the above expression in order to get the required velocity at the instant when 100 kg of sand is there in the cart. Let ${{m}_{0}}$ be the mass of the cart which is 150 kg and ‘t’ be the time at which the velocity is to be found. So, we can find the velocity as –
\begin{align} & F-\eta v=m\dfrac{dv}{dt} \\ & \text{but,} \\ & m={{m}_{0}}+\eta t \\ & \Rightarrow \dfrac{dm}{m}=\dfrac{dv}{F-\eta v} \\ & \Rightarrow \dfrac{dm}{{{m}_{0}}+\eta t}=\dfrac{dv}{F-\eta v} \\ & \text{Integrating w}\text{.r}\text{.t time on L}\text{.H}\text{.S and velocity on R}\text{.H}\text{.S} \\ & \Rightarrow \int\limits_{0}^{100}{\dfrac{dt}{{{m}_{0}}+\eta t}}=\int\limits_{0}^{v}{\dfrac{dv}{F-\eta v}} \\ & \Rightarrow \dfrac{1}{{{m}_{0}}}\int\limits_{0}^{t}{\dfrac{dt}{1+\dfrac{\eta t}{{{m}_{0}}}}}=\dfrac{1}{F}\int\limits_{0}^{v}{\dfrac{dv}{1-\dfrac{\eta v}{F}}} \\ & \Rightarrow \ln \left| 1+\dfrac{\eta t}{{{m}_{0}}} \right|=-\ln \left| 1-\dfrac{\eta v}{F} \right| \\ & \Rightarrow 1+\dfrac{\eta t}{{{m}_{0}}}=\dfrac{1}{1-\dfrac{\eta v}{F}} \\ & \Rightarrow v=\dfrac{Ft}{{{m}_{0}}+\eta t} \\ & \text{given,} \\ & F=10N \\ & {{m}_{0}}=150kg \\ & \eta =100\times {{10}^{-3}}kg{{s}^{-1}} \\ & m=100kg \\ & t=\dfrac{m}{\eta }=1000s \\ & \Rightarrow v=\dfrac{10\times 1000}{150+(100\times {{10}^{-3}})(1000)} \\ & \therefore v=40m{{s}^{-1}} \\ \end{align}

So, the velocity of the cart-sand system when there is 100 kg sand in the cart is 40 $m{{s}^{-1}}$.

Thus, the correct answer is option C.

Note:
The force applied to a body is responsible for any acceleration of the body. The acceleration is dependent on the mass. For a body of constant mass, the velocity increases linearly, but when the mass is changing the velocity will be a function of both mass and the time.