Question

Two football players collide head-on in midair while trying to catch a thrown football and cling together. The first player is 94 kg and has an initial velocity of 6 m/s, while the second player is 116 kg and has an initial velocity of -5 m/sn the next three parts, (d) through (f), justify that friction could be ignored compared to the forces of collision by considering the change in momentum of the first player. Let's assume that the collision lasts for 10ms. Calculate the force on the first player by the second player during the collision

Answer 1

Answer:

$F = 57116.2 N$

Explanation:

As we know that during collision the force between two players due to all external factors is very small or negligible

So we can use momentum conservation to find the final common speed

$m_1v_1 + m_2v_2 = (m_1 + m_2) v$

so we have

$94(6) + (116)(-5) = (94 + 116) v$

$v = -0.076 m/s$

now we can find acceleration which is defined as rate of change in velocity

So it is given as

$a = \frac{v_F - v_i}{\Delta t}$

$a = \frac{-0.076 - 6}{10 \times 10^{-3}}$

$a = - 607.62 m/s^2$

Now force on first player is given as

$F = ma$

$F = 94 \times (607.62)$

$F = 57116.2 N$