Question

Two football players collide head-on in midair, moving along the same horizontal direction, while trying to catch a thrown football. The first player is 89.5 kg and has an initial velocity of 6.05 m/s (in the positive direction), while the second player is 111 kg and has an initial velocity of –3.55 m/s. What is their velocity just after impact if they cling together?

Answer 1

Answer:

The velocity just after the impact is 0.722m/s.

Explanation:

The velocity after the impact if they cling together can be determined by means of the equation for the conservation of the linear momentum:

$Qb = Qa$  (1)

Where $Q_{b}$ is the total linear momentum of the system before the collision and $Q_{a}$ is the total linear momentum of the system after the collision:

Remember that the total linear momentum of a system is the sum of the momentum of each member ($p = m\cdot v$)

$m_{1}\cdot v_{1} + m_{2}\cdot v_{2} = m_{1}\cdot v_{1} + m_{2}\cdot v_{2}$  (2)

Where $m_{1}$ is the mass of the first player, $v_{1}$ is the velocity of the first player, $m_{2}$ is the mass of the second player and $v_{2}$ is the velocity of the second player.

Equation 2 establishes how the linear momentum is conserved in the system, if there is not external force acting on it or if the sum of the net force is zero.

To get the velocity just after the impact, equation 2 can be expressed in the following way:

$m_{1}\cdot v_{1} + m_{2}\cdot v_{2} = (m_{1} + m_{2})v$ (3)

Finally, v can be isolated from equation 3

$v = \frac{m_{1}\cdot v_{1} + m_{2}\cdot v_{2}}{(m_{1} + m_{2})}$

$v = \frac{(89Kg)\cdot (6.05m/s) + (111Kg)\cdot (-3.55m/s)}{(89Kg + 111Kg)}$

$v = 0.722m/s$

Hence, the velocity just after the impact is 0.722m/s.