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?!

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

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skyluke89 skyluke89 · Answer 2 · 2018-10-26T10:46:47+02:00

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:

[tex]p_1 = m_1 v_1 =(110 kg)(2.5 m/s)=275 kg m/s[/tex]

while the momentum of the second tackler is

[tex]p_2 = m_2 v_2 =(82 kg)(-5.0 m/s)=-410 kg m/s[/tex]

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

[tex]p_i = p_1 +p_2 =275 kg m/s -410 kg m/s=-135 kg m/s[/tex]

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

[tex]p_f = -135 kg m/s[/tex]

Which is also equal to

[tex]p_f = (m_1 +m_2 )v_f[/tex]

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:

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

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