Question

A flat cart of mass $m_0$ starts moving to the right due to a constant horizontal force $F$ at $t=0$. Sand spills on the flat cart from a stationary hopper. The velocity of loading is constant and is equal to $\mu \text{kg/s}$. Then,

• A. Initial acceleration is equal to $F/m_0$
• B. Acceleration at time $t$ is $\frac{F}{(m_0+\mu t)}$
• C. Kinetic energy of loaded cart at an instant is equal to work done by force $F$ up to that instant
• D. Momentum of loaded cart at an instant is equal to impulse of force $F$ up to that instant

Answer 1

Answer: (A), (D)

Momentum of loaded cart at an instant is equal to impulse of force $F$ up to that instant

Let at an instant ${}^{\prime }{t}^{\prime }$, velocity of cart be $v$. Due to the sand falling on the cart, it experiences a retarding force, which is equal to $\mu v$. It means net accelerating force on the cart is equal to $(F-\mu v)$. But at this instant, mass of the loaded cart is equal to $(m_0+\mu t)$.
Therefore, acceleration at time $t$ will be equal to $\frac{(F-\mu v)}{(m_0+\mu t)}$. It means option (b) is incorrect.
If in the above equation, $t$ is substituted as zero, then initial acceleration of the cart is
$a_0=\frac{(F-\mu v)}{m_0}$
But at initial moment, speed of flat cart is equal to zero. Hence, the initial acceleration becomes $\frac{F}{m_0}$. It means, option (a) is correct.
If we consider a system of flat cart and falling sand particles, then sand particles exert a retarding force on the cart which is equal to $\mu v$ and the cart exerts an accelerating force on sand particles which is also equal to $\mu v$. It means the only external force or the only resultant force acting on the system is equal to $F$.