Question: A force in the negative direction of an x axisis applied for 28 ms to

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Question:
A force in the negative direction of an x axisis applied for 28 ms to a 0.61 kg ball initially moving at 29 m/sin the positive direction of the axis. The force varies inmagnitude, and the impulse has magnitude 26.1 Ns. (a) What is the ball's velocity(including sign for direction) just after the force isapplied? (b) What is the averagemagnitude of the force on the ball?

Show solutionin detail so i can understand how it is solved. thank you for yourtime. Question:


Show solutionin detail so i can understand how it is solved. thank you for yourtime.

Explanation / Answer

a) let the speed be v after the force impulse = -26.1 N s = P = mv-m*29m/s =>v = (-26.1N/s / 0.61kg) + 29m/s = -13.8 m/s it is moving negative x- axis with speed -13.8m/s b) impulse = -26.1 N.s = F.t =>F = -26.1 N.s/t = -26.1N.s / 28*10-3 s =-932.14N the force is point to negative with magtitude F =932.14N

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