# On a touchdown attempt, a 95.0 kg running back runs toward the end zone at 3.75 m/s. A 111 kg linebacker moving at 4.10 m/s meets the runner in a head-on collision. If the two players stick together, what is their velocity immediately after the collision?

Oct 23, 2015

$v = 0.480 m . {s}^{- 1}$ in the direction that the linebacker was moving in.

#### Explanation:

The collision is inelastic as they stick together. Momentum is conserved, kinetic energy is not.

Work out the initial momentum, which will be equal to the final momentum and use that to solve for the final velocity.

Initial momentum.
Linebacker and runner are moving in opposite directions… choose a positive direction. I will take the direction of the linebacker as positive (he has larger mass and velocity, but you can take the runner's direction as positive if you want, just be consistent).
Terms : ${p}_{i}$, total initial momentum; ${p}_{l}$, linebacker's momentum; ${p}_{r}$, runner's momentum.

p_i = p_l + p_r = 111 × 4.10 + 95.0 × (-3.75) = 455.1 - 356.25 = 98.85 kg.m.s^(-1)
That is, $98.85 k g . m . {s}^{- 1}$ in the direction of the linebacker because the value is positive.

Apply conservation of momentum.
Total final momentum, ${p}_{f} = {p}_{i}$.
Runner and linebacker "stick" together, so their masses combine. After the collision there is only one object moving (i.e. linebacker + runner). So now :
p_f = m_(l+r) × v_(l+r) ⇒ v_(l+r) = p_f / m_(l+r)

${v}_{l + r} = \frac{98.85}{111 + 95} = 0.480 m . {s}^{- 1}$

The velocity is positive indicating that the two move in the direction that the linebacker was moving in.