Question

A 110 kg tackler moving at 2.5m/s meets head on (and holds on to) an 82 kg halfback moving at 5.0m/s. what will be their mutual speed immediately after the collision?!

Answer 1

The situation in the problem is called an inelastic collision where kinetic energy is not conserved only momentum is conserved. We do as follows:

m1v1 + m2v2 = (m1 + m2)vf

vf = (m1v1 + m2v2) / (m1 + m2)
vf = (110)(2.5) + (82)(5.0) / (110 + 82)
vf = 3.57 m/s

Answer 2

Answer:

0.7 m/s

Explanation:

We can solve the problem by using conservation of momentum.

Before the collision, the momentum of the first tackler is:

$p_1 = m_1 v_1 =(110 kg)(2.5 m/s)=275 kg m/s$

while the momentum of the second tackler is

$p_2 = m_2 v_2 =(82 kg)(-5.0 m/s)=-410 kg m/s$

Note that we used a negative sign because the direction of the second tackler is opposite to that of the first tackler.

Therefore, the total momentum before the collision is

$p_i = p_1 +p_2 =275 kg m/s -410 kg m/s=-135 kg m/s$

Since the total momentum is conserved, this is also equal to the final total momentum :

$p_f = -135 kg m/s$

Which is also equal to

$p_f = (m_1 +m_2 )v_f$

since the two tacklers continue their motion together with final velocity vf. Re-arranging the previous equation, we can find the the new velocity of the two tacklers:

$v_f = \frac{p_f}{m_1 +m_2}=\frac{-135 kg m/s}{110 kg + 82 kg}=-0.7 m/s$

and the negative sign means the direction is the one of the second tackler.