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a. A rail road car of mass M is moving without friction on a straight horizontal track with a velocity ct. A man of mass m lands on it normally from a helicopter. What will be the new velocity of the car? ltbgt b. If now the man begins to run on it with speed um with respect to car in a direction opposite to motion of the car, what will be the new velocity of the car? |
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Answer» Let `v_(1)` be the new velocity of the car. Then according to the principle of conservatioin of linear momentum `Mu=(M+m)v_(1)` `implies v_(1)=m/(M+m)u`…………..i b. Let `v_(2)` be the new velocity of the car relative to the earth, then velocity of the man relative of the earth `=(v_(2)-v_(rel))` `:. v_(2)=v_(1)+(mv_(rel))/(M+m)` |
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