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# Momentum, Impulse, and Collisions

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Starting at t= 0, a horizontal net force F = (0.280 N/s)ti + (-0.450 N/s^2)t^2j is applied to a box that has an initial momentum p= (-3.00 kg.m/s)i + (4.00 kg.m/s)j. What is the momentum of the box at t= 2.00s?

https://brainmass.com/physics/applied-physics/momentum-impulse-collisions-470412

## SOLUTION This solution is FREE courtesy of BrainMass!

Force F = 0.28 ti + (-0.45)t^2j
Initial momentum (@ t = 0) p_initial = 3.00i + 4.00j
Final momentum (@ t = 2.00 sec) p_final = ?

Force F = dp/dt
=> dp = F*dt
=> integration (dp) [p_initial to p_final] = integration (F*dt) [ t = 0 to t = 2]
=> p_final - p_initial = [(0.28*t^2/2)i + (-0.45*t^3/3)j] [t = 0 to t = 2]
=> p_final = (3.00i + 4.00j) + (0.14*4i - 0.15*8j)
=> p_final = 3.56i +2.80j kg-m/s

Therefore, the momentum of the box a t = 2 seconds, is 3.56i +2.80j kg-m/s.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!