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

Answer:

The velocity of cart B is $\sqrt{2}$ 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, $a_{A} = \frac{F}{m}$

Acceleration of the cart B, $a_{B} = \frac{2F}{m} = 2[tex][a_{A}]$

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

For cart A:

$v_{A}^{2} = u_{A}^{2} + 2a_{A}D$

$v_{A} = \sqrt{2aD}$

$v_{B}^{2} = u_{B}^{2} + 2a_{B}D$

$v_{B}^{2} = 2(2a_{A})}D$

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:

$\frac{v_{A}}{v_{B}} = \sqrt{\frac{1}{2}}$

$v_{B} = \sqrt{2}v_{A}$