Question

Carts A and B have equal masses and travel equal distances D on side-by-side straight frictionless tracks while a constant force F acts on A and a constant force 2F acts on B. Both carts start from rest. The velocities A and B of the bodies at the end of distance D are related by

Answer 1

The velocity of cart B is
`√(2)` the velocity of cart A

Solution:

As per the question:

Let the masses of both the carts A and B be 'm' kg

Distance traveled by both the carts be 'D' m

Force acting on A be 'F' N

Force acting on B be '2F' N

Now,

The relation between the velocities of A and B can be derived as :

Acceleration of cart A,
`a_(A) = (F)/(m)`

Acceleration of the cart B,
`a_(B) = (2F)/(m) = 2[tex][a_(A)]`

Now, using the third eqn of motion for both the carts A and B:

For cart A:

`v_(A)^(2) = u_(A)^(2) + 2a_(A)D`

`v_(A) = √(2aD)`

`v_(B)^(2) = u_(B)^(2) + 2a_(B)D`

`v_(B)^(2) = 2(2a_(A))}D`

where

`u_(A) = u_(B) 0` = initial velocity of cart A and cart B respectively

`v_(A)` = final velocity of cart A

`v_(B)` = final velocity of cart B

`v_(B) = \sqrt{4a_(A)}D`

Now, dividing the velocities of the cart A and B:

`(v_(A))/(v_(B)) = \sqrt{(1)/(2)}`

`v_(B) = √(2)v_(A)`