Question

A 200 kg cart is coasting at a constant velocity of 25 m/s. As the cart coasts 50 kg block is dropped into it. What will be the velocity of the cart now ? m/s

Answer 1

The final velocity of the cart, after the block is dropped into it, is 17.5 m/s.

Step-by-step explanation:

To solve this problem, we can use the conservation of momentum principle. The initial momentum of the cart can be calculated by multiplying its mass by its initial velocity. The initial momentum of the 50 kg block can be calculated by multiplying its mass by its initial velocity.

Since both the cart and the block are initially at rest, their initial velocities are 0 m/s.

After the block is dropped into the cart, both objects stick together and move with a final velocity. The final momentum of the combined system can be calculated by adding the initial momenta of the cart and the block. Finally, the final velocity of the combined system is the ratio of the final momentum to the combined mass of the cart and the block.

Using these calculations, the final velocity of the cart, after the block is dropped into it, is 17.5 m/s.

Answer 2

The initial kinetic energy of the cart is

`K= (1)/(2)Mv_(i)^2 = (1)/(2)\cdot 200 kg \cdot (25 m/s)^2=62500 J`
After the block is dropped into the cart, if there are no other forces acting on it, the kinetic energy of the new system cart+block must be conserved, so it should be the same as before. But the new mass will be M+m, where m=50kg is the mass of the block. Therefore we can write

`K= (1)/(2) (M+m) v_f^2`
from which we find

`v_f = \sqrt{ (2K)/(M+m) }= \sqrt{ (2\cdot 62500 J)/(200kg+50kg) }=22.36 m/s`