question_answer

An iron sphere of mass 1 kg is dropped from a height of 10 m. If the acceleration of sphere is 9.8 \[\text{m}{{\text{s}}^{-2}}\], calculate the momentum transferred to the ground by the ball.

Answer:

                               Here, initial velocity of sphere, u = 0 Distance travelled, s = 10 m Acceleration of sphere, a = 9.8 \[\text{m}{{\text{s}}^{-2}}\] Final velocity of sphere when it just reaches the ground can be calculated using \[{{\upsilon }^{2}}-{{u}^{2}}=2as\]or\[{{\upsilon }^{2}}-0=2\times 9.8\times 10=196\,\text{m/s}\] or\[\upsilon =\sqrt{196}m/s=14m/s\]                    Momentum of the sphere just before it touches the ground = mv \[=1kg\times 14kg-m{{s}^{-1}}\]                                               \[=14kg-m{{s}^{-1}}\] On reaching the ground, the iron sphere comes to rest, so its final momentum is equal to zero according to the law of conservation of momentum. Momentum transferred to the ground = Momentum of the sphere just before it comes to rest = 14 kg -\[\text{m}{{\text{s}}^{\text{-1}}}\]