Irodov Problem 1.156

a) The total momentum of the (buggy + two men) system will be conserved just before and after their jump. Initially everyone is at rest and so the total momentum is 0. Suppose that after the men jump the buggy moves with a velocity v, then its momentum will be Mv. Since the men jump into the air with a speed u relative to the buggy, their speed in air with respect to a stationary observer will be v+u, and so their combined momentum in air will be 2m(v+u). Hence we have,







(b) We have to solve this problem in two parts, first after the first man jumps and then after the second man jumps. Suppose that after the first man jumps, the buggy attains a velocity v1 and then suppose that after the second man jumps it attains a velocity v2.

Just before the first man jumps off the system is at rest and so the total momentum is 0. Just after the first man jumps, the momentum of buggy and one remaining person will be (M+m)v1. The man jumps with a speed u relative to the buggy and so his speed in the air relative to a stationary reference will be m(v1+u). Since, there are no external forces, the momentum of the (buggy + 2 men) system just before and after the first man jumps will be conserved. So we have,






Now the second man jumps. Before he jumps the momentum of the (buggy + remaining man) is given by (M+m)v1. After the man jumps, the momentum of the buggy becomes Mv2. Since the man jumps with a speed u relative to the buggy his speed in air relative to a stationary observer is v2+u and his momentum in air m(v2+u). Since there are no external forces on the (remaining man + buggy) system before and after his jump, the momentum of the (remaining man + buggy) system must be conserved. So we have,

Irodov Problem 1.155

Suppose that the final velocities of the rear and front buggies after jump became v1 and v2 respectively.

For the rear buggy, momentum before the jump is (M+m)v0. After the jump the rear buggy moves at v1 and so with a momentum Mv1. The person in the air moves with a velocity u relative to the rear buggy and so his velocity relative to a stationary observer is v1 + u. The momentum of the person in the air is thus m(v1+u). Since, there was no external force acting on the (rear buggy + man) system their total momentum must be conserved immediately before and after jump. This means that,






For the front buggy, the momentum of the (man + buggy) system just before and after he lands is conserved. Before he lands into the front buggy his momentum in the air is m(v1 + u) and the momentum of the buggy in front is Mv0 . After her lands, both the buggy and man move at a speed v2 together and so their combined momentum is (M+m)v2. In other words,

Irodov Problem 1.154

Essentially in this problem, the momentum of each buggy is conserved in their direction of motion, since there are no external forces in this direction acting on the buggies. Initially (before the people do the exchange jump) the momentum of the buggies are (m + M) v1 and (m+M)v2 respectively. When the people jump across their buggies, they do not loose their respective momenta in the air and so when they land in the opposite buggy, they add their respective momenta to the opposite buggy.

After the jump, for buggy 1 , the total momentum is the sum of the momentum of the buggy and the momentum of the man in buggy 2 (since he jumped into buggy 1), i.e. Mv1 + mv2. The momentum for buggy 2 after the jump is, the sum of the momentum of buggy 2 and the man in buggy 1 (since now he has jumped into buggy 2), i.e. Mv2 + mv1. The problem tells us that finally the buggy 1 stopped while buggy 2 now moves at a velocity v. This means that,



Solving these two equations we have,