Question

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.

Answer 1

Answer:

Explanation:

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

Answer 2

The momentum of the box at t = 1.90 s  is  - 2.586 i + 2.85j

And, the x and y component is - 2.586 and 2.85

How to calculate momentum?

Since we know that

impulse = force * time

Here impulse means the change in momentum

Now we can write as  in different way like

change in momentum = force * time

Also, Force = F = .285 t -.46t²

Due to variable force

change in momentum = ∫ F dt

So,

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

Sine t = 1.9

So,

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

= .514i - 1.05 j

And,

final momentum

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

= - 2.586 i + 2.85j

So, finally

x component = - 2.586

y component = 2.85

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