Question

Starting at t = 0 s , a horizontal net force F⃗ =( 0.290 N/s )ti^+(-0.440 N/s2 )t2j^ is applied to a box that has an initial momentum p⃗ = ( -2.85 kg⋅m/s )i^+( 4.00 kg⋅m/s )j^ . Part A 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

F= dp/dt

p = $\int$ Fdt

=> p_f - p_i = $\int$ [( 0.290 N/s )t i+(-0.440 N/s² )t² j] dt

=> p_f - (( -2.85 kg⋅m/s ) i+( 4.00 kg⋅m/s ) j) = [( 0.290 N/s )t² /2 i+(-0.440 N/s² )t³ /3 j]

For t= 1.90 s

=> p_f - (( -2.85) i+( 4.00) j) = [( 0.290* (1.90)² /2 i+(-0.440)*(1.90)³ /3 j]

=> p_f = [( 0.290* (1.90)² /2 i+(-0.440)*(1.90)³ /3 j] + (( -2.85) i+( 4.00) j

=> p_f = -2.33 kg⋅m/s i + 2.99 kg⋅m/s j

s0, p_f = -2.33 , 2.99