Starting at t = 0 s, a horizontal net force F⃗ = (0.285 N/s) ti^ + (-0.460 N/s2) t2j^ is applied to a box that has an initial momentum p⃗ = (-3.10 kg⋅m/s) i^ + (3.90 kg⋅m/s) j^.

What is the momentum of the box at t = 1.90 s?

Enter the x and y components of the momentum separated by a comma.

Question

Yoselin Herman

Physics

9 April, 14:13

Starting at t = 0 s, a horizontal net force F⃗ = (0.285 N/s) ti^ + (-0.460 N/s2) t2j^ is applied to a box that has an initial momentum p⃗ = (-3.10 kg⋅m/s) i^ + (3.90 kg⋅m/s) j^.

What is the momentum of the box at t = 1.90 s?

Enter the x and y components of the momentum separated by a comma.

+4

Answers (1)

Know the Answer?

Wyatt Rose · Answer 1

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F =.285 t -.46t²

Since force is variable

change in momentum = ∫ F dt where F is force

= ∫.285ti -.46t²j dt

=.285 t² / 2i -.46 t³ / 3 j

When t = 1.9

change in momentum =.285 x 1.9² / 2 i -.46 x 1.9³ / 3 j

=.514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

Starting at t = 0 s, a horizontal net force F⃗ = (0.285 N/s) ti^ + (-0.460 N/s2) t2j^ is applied to a box that has an initial momentum p⃗ = (-3.10 kg⋅m/s) i^ + (3.90 kg⋅m/s) j^.What is the momentum of the box at t = 1.90 s?Enter the x and y components of the momentum separated by a comma.