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Particle under F = 12t i − 24t² j − 40t³ k

Short answer

a_y = –12t², so at t=4 s, a_y = –192 m/s². Integrating from rest: v_y = –4t³ = –256 m/s, y = –t⁴ = –256 m. Principle of impulse and momentum: the impulse equals change in linear momentum, ∫F dt = mΔv.

The resultant external force acting on a 2 kg particle in space is F = (12t i − 24t² j − 40t³ k) N, where t is the time measured in seconds. The particle is at rest at the origin when t = 0. Determine the acceleration component ay, the velocity component Vy, and the coordinate y of the particle at the instant of 4 sec. What do you mean by principle of impulse and momentum?

Equations of motion (m = 2 kg)

F = 12t i − 24t² j − 40t³ k (N). From F = ma:

  • a_y = F_y/m = −24t²/2 = −12t² m/s²
  • At t = 4 s: a_y = −12(16) = −192 m/s²

Velocity (starts from rest)

v_y = ∫a_y dt = −4t³ At t = 4 s: v_y = −4(64) = −256 m/s

Coordinate (starts at the origin)

y = ∫v_y dt = −t⁴ At t = 4 s: y = −256 m

(The negative signs mean that the particle moves in the −y direction.)

Principle of impulse and momentum

Starting from Newton's second law, F = m dv/dt, multiply by dt and integrate between times t₁ and t₂:

∫ F dt = m v₂ − m v₁

  • Linear impulse = ∫ F dt (units N·s): the effect of a force acting over a time interval.
  • Linear momentum = m v (units kg·m/s).
  • The principle states: the linear impulse of the resultant force on a particle during a time interval equals the change in its linear momentum. In vector form m v₁ + ∫ F dt = m v₂, with separate component equations along x, y, z.
  • If the resultant force is zero, momentum is conserved. The principle is most useful when the force is given as a function of time or when velocity and time (not displacement) are involved, for example impact.

✓ Answer checked by Claude Opus 5.5 (high effort)

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