Question

Suppose a cart is moving without any loss of energy to friction or air resistance. As the cart passes under a bridge, a heavy block is dropped vertically downward into the cart from a very small height above the cart. What happens to the speed of the cart

Answer 1

Answer:

speed of the cart decreases

Explanation:

According to the conservation of momentum, as no external force is applied the linear momentum is conserved.

momentum before dropping = momentum after dropping

So, as the mass is dropped, the total mass is increased and thus to keep the momentum constant, the velocity of the cart decreases.

Thus, the speed of the cart decreases.

Answer 2

Answer:

The speed of the cart will decrease.

Explanation:

In the said system energy shall remain conserved as it is an isolated system.

Initially the whole kinetic energy was with the mass of the cart now when we add some of the mass to the cart it's mass will increase but the energy will be same as that of initially of the lighter cart thus it's speed decrease to keep the kinetic energy constant.

Mathematically

$K.E_{1}=\frac{1}{2}m_{cart}v^{2}\\\\K.E_{2}=\frac{1}{2}(m_{cart}+m_{block})v_{2}^{2}\\\\\therefore K.E_{1}=K.E_{2}\\\\\Leftrightarrow v_{2}=\sqrt{\frac{m_{cart}}{(m_{cart}+m_{block})}}v\\\\\therefore v_{2}< v$