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 mom
entum 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.
1 answer:
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
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