Question

A 91.0-kg fullback running east with a speed of 5.20 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s.

(a) Explain why the successful tackle constitutes a perfectly inelastic collision.

(b) Calculate the velocity of the players immediately after the tackle.

magnitude	m/s
direction	° north of east

(c) Determine the mechanical energy that disappears as a result of the collision.
J

(d) Account for the missing energy.

Answer 1

There are two reasons: First, momentum is always conserved in a closed system (which, for all practical purposes, we can assume this example is). Secondly, momentum is a vector whereas KE is not; we can use the direction and magnitude of the players' original momenta to figure out how fast they'll be moving post-collision.

So let's start with the fullback. Momentum = mass x velocity:

p = mv
p = (91 kg)(5.20 m/s) = 473.2 kg*m/s East

Since we'll need it later, let's go ahead and find his KE as well:

KE = 1/2 mv²
KE = 1/2 (91 kg)(5.20 m/s)²
KE = (45.5 kg)(27.04 m²/s²)
KE = 1230.32 J

Now the opponent's momentum and KE:

p = mv
p = (95 kg)(3.0 m/s) = 290 kg*m/s North

KE = 1/2 mv²
KE = 1/2 (95 kg)(3.0 m/s)²
KE = (48 kg)(9.0 m²/s²)
KE = 430 J

So the players' total KE prior to the collision is 1230.32 J + 430 J = 1660.32 J. We'll come back to that later.

After the collision, the momentum of the players will be neither due North nor East, but somewhere in between. In order to find that momentum (so we can find their velocity), we need to use vector addition.

Two vectors to add:
473.2 kg*m/s East
290 kg*m/s North

The hypotenuse of a right triangle formed by placing those vectors head-to-tail is: 540 kg*m/s. The angle is really unimportant since we know the magnitude (and that's all we need to find the players' speed).

The combined mass of the players is 91 kg + 95 kg = 186 kg. So, the velocity of the players following the collision is:

v = p/m = (540 kg*m/s) / (186 kg) = 2.9 m/s

Now that we have the velocity of the players, we can find their total KE following the collision:

KE = 1/2 mv²
KE = 1/2(186 kg)(2.9 m/s)²
KE = 782.13 J

Subtracting from the original KE:

1660.32 J - 782.13 J = 878.19 J