Question

16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both players end up with zero speed.
a) What was the fullback’s momentum before the collision?

b) what was the change in the fullback’s momentum?

c) what was the tackle’s original momentum?

e) how fast was the tackle moving originally?

Answer 1

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

$p=mv=(95)(8.2)=779 kg m/s$

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

$\Delta p = p' -p =0-779 = -779 kg m/s$

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

$p_f + p_t = p'$

where

$p_f$ is the initial momentum of the fullback

$p_t$ is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

$p_f = 779 kg m/s\\p' = 0$

Therefore, we can find the tackle's original momentum:

$p_t = p'-p_f = 0-(779) = -779 kg m/s$

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

$p=-779 kg m/s$ is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

$v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s$

So, he was moving at 6.1 m/s in the direction opposite to the fullback.