Calculate the initial velocities of buggies

[utkarshakash]

Homework Statement

Two identical buggies 1 and 2 with one man in each move without friction due to inertia along the parallel rails toward each other. When the buggies get opposite each other, the men exchange their places by jumping in the direction perpendicular to the motion direction. As a consequence, buggy I stops and buggy 2 keeps moving in the same direction, with its velocity becoming equal to v. Find the initial velocities of the buggies v1 and v2 if the mass of each buggy (without a man) equals M and the mass of each man m.

Homework Equations

Momentum Conservation

The Attempt at a Solution

Let the buggy 2 move in the -ve X direction and buggy 1 in +ve X.
Along the direction of motion momentum conservation principle can be applied.
$P_{initial}=(m+M)v_{1}-(M+m)v_{2}$
$P_{final}=-(M+m)v$
From this I get a relation between three
$v_{2}=v_{1}+v$

What to do next?

 

[anathema]

Thank you for your question. After finding the relation between v1 and v2, we can use the conservation of momentum equation in the perpendicular direction to find the initial velocities.

Let's define the positive direction as the direction in which buggy 1 is initially moving. Then, the momentum conservation equation in the perpendicular direction can be written as:

(m+M)v1 = mv2

Substituting v2 = v1 + v, we get:

(m+M)v1 = m(v1 + v)

Solving for v1, we get:

v1 = v(m+M)/(m+2M)

Similarly, using the same equation in the perpendicular direction for buggy 2, we get:

(m+M)v2 = Mv1

Substituting the value of v1 from the previous equation, we get:

v2 = vM/(m+2M)

Therefore, the initial velocities of buggy 1 and buggy 2 are:

v1 = v(m+M)/(m+2M)

v2 = vM/(m+2M)

I hope this helps. Let me know if you have any further questions.