Name: Two Block System
Uploaded: 2019-07-19
Duration: 6 min 20 s
Description: Two blocks placed on a surface exerts a force between them and also with the surface. Let us discuss how these forces can be solved numerically in this video.

Question

A cart of total mass $M_{0}$ is at rest on a rough horizontal road. It ejects bullets at rate of $λ kg s^{- 1}$ at an angle $θ$ with the horizontal and at velocity $u$ (constant) relative to the cart. The coefficient of friction between the cart and the ground is $μ$ . Find the velocity of the cart in terms of time $t$ . The cart moves with sliding.

Accepted Answer

When the cart will eject the bullet, then a thrust force will be applied to the bullet which is given by,

$F_{T} = v_{rel} \frac{d m}{d t} = λ v_{rel}$

The mass of the cart after time $t$ will be,

$m = M_{0} - λ t . . . (1)$

From the free body diagram,

$F_{net} = F_{T_{x}} - f$

$m a = u \cos θ - μ N . . . (2)$

$N = m g + F_{T_{y}}$

$N = m g + λ u \sin θ . . . (3)$

From equation $(1)$ , $(2)$ and $(3)$ ,

$a = \frac{d v}{d t} = \frac{λ u (\cos θ - μ \sin θ)}{(M_{0} - λ t)} - μ g$

$\int_{0}^{v} d v = \int_{0}^{t} \frac{λ u (\cos θ - μ \sin θ)}{(M_{0} - λ t)} d t - μ g \int_{0}^{t} d t$

$v = - λ u (\cos θ - μ \sin θ) \ln {[M_{0} - λ t]}_{0}^{t} - μ g t$

$v = λ u (\cos θ - μ \sin θ) \ln (\frac{M_{0}}{M_{0} - λ t}) - μ g t$ .

Important Questions on Friction

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