Answer:
v = 0.16 m/s in the positive x direction.
Explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved.
- Initial momentum can be expressed as follows:
[tex]p_{o} =m_{1} * v_{o1} + m_{2} * v_{o2} (1)[/tex]
- Since both players move only horizontally, (1) can be expressed in terms of the velocities in the x- directions only, taking into account that player 1 moves in the positive x-direction, and player 2 in the opposite direction, and assuming that positive velocities are in the positive x-direction also:
[tex]p_{o} =110 kg * 4.4 m/s + 135 kg * (-3.3 m/s) (2)[/tex]
- Since total momentum must be conserved, the final momentum must be equal to the initial one.
- We know that both players cling together after colliding, so the collision is totally inelastic.
- So, the final momentum can be expressed as follows:
[tex]p_{f} =( m_{1} + m_{2} )* v_{f} (3)[/tex]
- Replacing by the givens in (3), we get:
[tex]p_{f} =( m_{1} + m_{2} )* v_{f} = (110 kg + 135 kg)* vf (4)[/tex]
- Since (3) must be equal to (4) we can solve for vf, as follows:
[tex]v_{f} = \frac {110 kg * 4.4 m/s + 135 kg * (-3.3 m/s) }{(110 kg + 135 kg)} = 0.16 m/s (5)[/tex]
- Since the initial momentum has only component in the x-direction, the final one must have component in the x-direction only too.
- This means that vf is in the x-direction, and due to it is positive, it must aim in the positive x-direction.
- So, vfx = +0.16 m/s
