Solution: The second Newton's law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force. "\\vec F=\\frac{d\\vec p}{dt}" In our task the directions of force and momentum coincides, and we can write for average magnitudes:
"F=\\frac{\\Delta p}{\\Delta t}" , where increment of the momentum is "\\Delta p= m\\cdot \\Delta v" . Since the ball is initial at rest the speed increment coincides with the final speed "v" of the ball. We have
"F\\cdot \\Delta t=m \\cdot v" , and "v=\\frac {F\\cdot \\Delta t}{m}=\\frac{60N\\cdot 10\\cdot 10^{-3}s}{0.25 kg}=2.4 ms^{-1}"
Answer: speed of the ball just after the impact is 2.4m/s
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