Question

A cart loaded with sand moving with velocity \[V\] sand is falling through the hole as shown in diagram then after falling on the ground sand is:

A. Moving with cart
B. Moving in opposite direction with speed $v$
C. Stationary on the ground
D. Moving with speed ${v_2}$

Answer 1

Hint: We must determine the direction and speed of the sand in order to answer the question. To address this, we'll apply the notion of conservation of momentum, but first, let's get a fundamental understanding of what it means. The law of conservation of momentum asserts that in a closed system, the overall momentum does not change.

Complete step by step answer:
Let us first write all the given values accordingly; the velocity of a sand-filled cart is equal to \[v\]. Now, let us consider the mass of the cart be ${m_1}$ , mass of the sand be ${m_2}$ and velocity of the sand be ${v_2}$.Here we need to find after falling on the ground, the direction and speed of the sand and for that we will apply the law of conservation of momentum principle.

Now, according to the law of conservation of momentum,
\[\left( {{m_1} + {m_2}} \right)v = {m_1} \times v + {m_2} \times {v_2} \\
\Rightarrow {m_1}v + {m_2}v = {m_1}v + {m_2}{v_2} \]
Equating further \[{m_1}v\] will be cancelled out on the both side
${m_2}v = {m_2}{v_2}$
Now, ${m_2}$ will cancelled out so the final value is
$v = {v_2}$
As a result, sand moves in the same direction and at the same speed as the cart.

So, the correct option is A.

Note: Individual bodies are not subject to the law of conservation of momentum, which applies to a system of particles. It's vital to remember that the system's momentum is preserved, not the individual particles'. Individual bodies' momentum can change depending on the situation, but the system's momentum will always be preserved as long as no external net force is exerted on it.