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

Their velocity afterwards is 2.88 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:

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

where: in this case:

$m_1 = 91.5 kg$ is the mass of the first player

$u_1 = 2.73 m/s$ is the initial velocity of the first player (choosing east as positive direction)

$m_2 = 63.5 kg$ is the mass of the second player

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

$v$ is their combined velocity afterwards

Solving for v, we find:

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

And the sign is positive, so the direction is east.

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