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?

Explanation

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)`