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
∑mi(vi)1+∑∫t2t1Fidt=∑mi(vi)2
where mi is the ith particle's mass, vi is the ith particle's velocity, and Fi is the external force that acts on the ith particle. This relationship states that the sum of the initial linear momenta, at time t1, and the impulses of all the external forces acting between times t1 and t2 is equal to the sum of the linear momenta of the system, at time t2. If the system has a mass center, G, the expression becomes
m(vG)1+∑∫t2t1Fidt=m(vG)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 μk = 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.
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 4 images