Question

**Principle of Linear Impulse and Momentum for a System of Particles**

To apply the principle of linear impulse and momentum to a system of particles.

Integrating the equation of motion, as applied to all particles in a system, yields

∑*m**i*(**v***i*)1+∑∫*t*2*t*1**F***i**dt*=∑*m**i*(**v***i*)2

where *m**i* is the *i*th particle's mass, **v***i* is the *i*th particle's velocity, and **F***i* is the external force that acts on the *i*th particle. This relationship states that the sum of the initial linear momenta, at time *t*1, and the impulses of all the external forces acting between times *t*1 and *t*2 is equal to the sum of the linear momenta of the system, at time *t*2. If the system has a mass center, *G*, the expression becomes

*m*(**v***G*)1+∑∫*t*2*t*1**F***i**dt*=*m*(**v***G*)2

This expression allows the principle of linear impulse and momentum to be applied to a system of particles that is represented as a single particle.

Two blocks, each of mass *m* = 6.90 kg , are connected by a massless rope and start sliding down a slope of incline *θ* = 36.0 ∘ at *t*=0.000 s. The slope's top portion is a rough surface whose coefficient of kinetic friction is *μ*k = 0.250. At a distance *d* = 1.40 m from block *A*'s initial position the slope becomes frictionless. (Figure 1)What is the velocity of the blocks when block *A* reaches this frictional transition point? Assume that the blocks' width is negligible.

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