Impulse-momentum principle; blocks on incline

Impulse–momentum principle

The linear impulse of the resultant force acting on a particle during a time interval is equal to the change in its linear momentum:

∫ F dt = m v₂ − m v₁ (or m v₁ + ∫ F dt = m v₂)

Linear impulse has units N·s and equals the area under the force–time graph. If the resultant force is zero the momentum is conserved. For a constant force, F t = m(v₂ − v₁), which is Newton's second law in integrated form.

Blocks A and B

A (500 N) lies on a 15° incline and is joined by a cord over a frictionless, massless pulley to B (1000 N) hanging vertically. B is heavier, so B descends and A is pulled up the incline. μ_k = 0.35. Let a be the common acceleration and T the cord tension.

• Normal reaction on A: N = 500 cos 15° = 482.96 N
• Friction on A (down the incline): F = 0.35 × 482.96 = 169.04 N
• Component of A's weight along the incline: 500 sin 15° = 129.41 N

   A \ (500 N)        15° incline, A moves up
      \ ___ o pulley
              |
              B (1000 N) moves down

Block B: 1000 − T = (1000/9.81) a Block A: T − 129.41 − 169.04 = (500/9.81) a

Adding: 1000 − 298.45 = (1500/9.81) a 701.55 = 152.91 a

a = 4.59 m/s² for both blocks (A up the slope, B down).

Tension: T = 1000 − (1000/9.81)(4.588) = 532.3 N

Answer: a_A = a_B = 4.59 m/s²; T = 532.3 N.