Question

A block of mass 5 kg is at rest on a smooth horizontal surface. Water coming out of a pipe horizontally at the rate of ${\rm{2}}\;{\rm{kg - }}{{\rm{s}}^{ - 1}}$, hits the block with a velocity of ${\rm{6}}\;{\rm{m - }}{{\rm{s}}^{ - 1}}$. The initial acceleration of the block is,
(1) Zero
(2) $1.2{\rm{ m}}{{\rm{s}}^{ - 2}}$
(3) $2.4{\rm{ m}}{{\rm{s}}^{ - 2}}$
(4) $0.6{\rm{ m}}{{\rm{s}}^{ - 2}}$

Answer 1

In the given question, we need to apply the concept of force and momentum. The force can be described as the rate of change of momentum. The following description means that not only the change of velocity will affect the force applied but also the change of mass (rather the rate of change of mass) will affect the value of force.

Complete step by step answer:
Given;
The mass of the block is 5 kg.
According to the question, the block rests on the smooth horizontal surface until it is hit by the water coming out of the pipe horizontally. So, the water hitting the block must impart some force on the block otherwise, there is no other force to give the block an acceleration.
Since the force can be defined as the rate of change of momentum, then,

F = \dfrac{\partial }{{\partial t}}\left( {mv} \right)\\\ F = \dfrac{{\partial m}}{{\partial t}} \times v + m \times \dfrac{{\partial v}}{{\partial t}} \end{array}$$ As, there is no change of velocity mentioned in the question, so $$\dfrac{{\partial v}}{{\partial t}} = 0$$. Also, the water is flowing out of the pipe at a specific rate, so there is a change in the amount of mass of water flowing out per unit time. We can say that, rate of change of mass is equal to rate at which the water flows out of the pipe On substituting all values in the former expression of force, we have, $$\begin{array}{l} F = \dfrac{{\partial m}}{{\partial t}} \times v\\\ F = 2{\rm{ kg}}{{\rm{s}}^{ - 1}} \times 6\;{\rm{m}}{{\rm{s}}^{ - 1}}\\\ F = 12\;{\rm{N}} \end{array}$$ Also, from Newton’s second law of motion, we have, $$F = m \cdot a$$ So, substitute the value of force obtained in the following expression we get; $$\begin{array}{l} \;\;\;\;\;\;F = m \cdot a\\\ 12 = 5\;{\rm{kg}} \cdot a\\\ {\rm{ }}\;a = 2.4\;{\rm{m}}{{\rm{s}}^{ - 2}} \end{array}$$ Therefore, the initial acceleration of the block is $$a = 2.4\;{\rm{m}}{{\rm{s}}^{ - 2}}$$, and option (3) is correct. **Note:** In the given question, we have used two concepts of force: the rate of change of mass and the other as the rate of change of velocity. This concept is a derivation from the concept that force is the rate of change of momentum; if we put either the rate of change of mass to be zero or the rate of change of velocity to be zero, we will end up in either of the former conclusions.