Question

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity afterward? PLEASE HELP

Answer 1

Answer:

The velocity of the players will be 2.88 m/s in the east direction.

Explanation:

Let 'v' be the velocity of the players after collision.

Consider the east direction as positive direction.

Given:

Mass of the first player  is, $m_1 = 91.5$ kg  

Initial velocity of the first player  is, $u_1 = 2.73$ m/s  

Mass of the second player  is, $m_2 = 63.5$ kg  

Initial velocity of the second player is, $u_2 = 3.09$ m/s  

In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

Solving for v, we get:

$v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88\ m/s$

Therefore, their velocity after the collision is 2.88 m/s.

The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.