Question

A 95 kg fullback is running at 3.0 m/s to the east and is stopped in 0.85 s by a head-on tackle by a tackler running due west. Calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler, and (d) the average force exerted on the tackler.

Answer 1

(a) Original momentum of the fullback: 285 kg m/s

The original momentum of the fullback is equal to the product between his velocity and its mass:

$p_i=mv$

where:

m = 95 kg is the mass

v = 3.0 m/s is the velocity

Substituting the numbers into the formula, we find

$p_i=(95 kg)(3.0 m/s)=285 kg m/s$

(b) Impulse exerted on the fullback: -285 kg m/s

The impulse exerted on the fullback is equal to his variation of momentum:

$I=\Delta p=p_f -p_i$

where:

$p_f = 0$ is the final momentum of the fullback (zero because he comes to a stop)

$p_i = 285 kg m/s$ is the initial momentum

Substituting,

$I=0-(285 kg m/s)=-285 kg m/s$

(c) Impulse exerted on the tackler: 285 kg m/s

The total momentum of the fullback and the tackler must be conserved:

$p_i + P_i = p_f + P_f$

where $p_i$ and $p_f$ are the initial and final momentum of the fullback, while $P_i, P_f$ are the initial and final momentum of the tackler.

We can re-arrange the equation as follows:

$P_f - P_i = p_i -p_f\\\Delta p_{tackler} = -\Delta p_{fullback}\\I_{tackler} = -I_{fullback}$

which means that the impulse exerted on the tackler is the negative of the impulse exerted on the fullback, so:

$I=-(-285 kg m/s)=285 kg m/s$

(d) Average force exerted on the tackler: 335.3 N

The impulse on the tackler is equal to the product between the average force and the time of the collision:

$I=F \Delta t$

Since $\Delta t=0.85 s$, we can find the average force:

$F=\frac{I}{\Delta t}=\frac{285 kg m/s}{0.85 s}=335.3 N$