Question

Starting at ttt_1 = 0 ss , a horizontal net force F⃗ =(F→=( 0.265 N/sN/s )ti^+)ti^+(-0.460 N/s2N/s2 )t2j^)t2j^ is applied to a box that has an initial momentum p⃗ =p→= (( -3.00 kg⋅m/skg⋅m/s )i^+()i^+( 4.05 kg⋅m/skg⋅m/s )j^)j^ .

Answer 1

Answer:

The momentum of the box at t = 2 s is given as

p₂ = (-2.47î + 2.82ĵ) kgm/s

Explanation:

F = [(0.265 N/s)t] î + [(-0.460 N/s²)t²] ĵ

F = (0.265t) î - (0.460t²) ĵ

At t = 0

Initial momentum

p₀ = (-3.00 kg⋅m/s )î + (4.05 kg⋅m/s) ĵ

Find the momentum at t = 2 s

According to Newton's second law of motion, the change in momentum is equal to the impulse.

That is,

dp = F.dt

Δp = ∫²₀ F.dt

∫²₀ F.dt = ∫²₀ [(0.265t) î - (0.460t²) ĵ] dt

= [(0.1325t²) î - (0.1533t³) ĵ]²₀

= [0.1325(2²)] î - [0.1533(2³)] ĵ

∫²₀ F.dt= (0.53 î - 1.2267 ĵ) kgm/s

p₂ - p₀ = Δp = ∫²₀ F.dt = (0.53 î - 1.2267 ĵ)

p₂ = p₀ + Δp

p₂ = (-3.00î + 4.05ĵ) + (0.53 î - 1.2267 ĵ)

p₂ = (-2.47î + 2.8233ĵ) kgm/s

p₂ = (-2.47î + 2.82ĵ) kgm/s

Hope this Helps!!!