Question

Two football players collide head-on in midair while chasing a pass. The first player has a 110 kg mass and an initial velocity of 4.40 m/s in the positive x direction, while the second player has a 135 kg mass and initial velocity of 3.30 m/s in the negative x direction.
Required:
What is the x component of their velocity just after impact if they cling together?

Answer 1

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:

       $p_{o} =m_{1} * v_{o1} + m_{2} * v_{o2} (1)$

• 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:

       $p_{o} =110 kg * 4.4 m/s + 135 kg * (-3.3 m/s) (2)$

• 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:

       $p_{f} =( m_{1} + m_{2} )* v_{f} (3)$

• Replacing by the givens in (3), we get:

       $p_{f} =( m_{1} + m_{2} )* v_{f} = (110 kg + 135 kg)* vf (4)$

• Since (3) must be equal to (4) we can solve for vf, as follows:

       $v_{f} = \frac {110 kg * 4.4 m/s + 135 kg * (-3.3 m/s) }{(110 kg + 135 kg)} = 0.16 m/s (5)$

• 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