Question

Two football players are running along perpendicular paths. One player is moving to the right with a speed of 4 m/s and a mass of 100 kg. The other player is moving forward with a speed of 6 m/s and a mass of 50 kg. The smaller player tackles the larger one at the point of their collision and stay stuck together (this player actually knows how to wrap their arms when tackling), sliding along the ground. Take the coefficient of kinetic friction to be 0.20 for this problem. Find the following: a) the total momentum of the system just before the collision (magnitude and direction). b) the total momentum of the system just after the collision (magnitude and direction). c) the final velocity of the players (magnitude and direction) right after the collision. d) the kinetic energy of the system just after the collision. e) the distance the players slide before coming to rest

Answer 1

The total momentum before and after the collision of the two football players is 500 kg·m/s, and their final velocity is 3.33 m/s. The kinetic energy after collision is 833.25 J, and the distance they slide before coming to a rest is 2.83 meters.

Step-by-step explanation:

Let's calculate the properties of the system both pre-collision and post-collision for the two football players.

a) Total momentum before the collision

Momentum is a vector quantity, given by the product of mass and velocity. For the first player moving to the right (assuming right is the positive x-direction):
p1 = m1 * v1x = 100 kg * 4 m/s = 400 kg·m/s (in the +x direction).
For the second player moving forward (assuming forward is the positive y-direction):
p2 = m2 * v2y = 50 kg * 6 m/s = 300 kg·m/s (in the +y direction).
To find the total momentum, use the Pythagorean theorem for the magnitude:
Ptotal = sqrt(p1x^2 + p2y^2) = sqrt(400^2 + 300^2) = sqrt(250000) = 500 kg·m/s.
The direction can be found using the arctangent function.
b) Total momentum after the collision

According to the conservation of momentum, the total momentum just after the collision is the same as before the collision, since no external forces are involved. So, it remains 500 kg·m/s in both magnitude and direction.

c) Final velocity after the collision

The combined mass after the collision is m1 + m2 = 100 kg + 50 kg = 150 kg. Thus, the final velocity Vf can be found by dividing the total momentum by the combined mass:
Vf = Ptotal / (m1 + m2) = 500 kg·m/s / 150 kg = 3.33 m/s

d) Kinetic energy after the collision

The kinetic energy K is given by the formula K = 0.5 * m * Vf^2.
K = 0.5 * 150 kg * (3.33 m/s)^2 = 0.5 * 150 * 11.09 = 833.25 J.

e) Distance the players slide before coming to rest

To calculate the distance d that the players slide before coming to rest, we use the work-energy principle where the work done by friction is equal to the loss of kinetic energy. The force of kinetic friction is µk * m * g, so work done W by friction is W = * d. Setting W equal to the kinetic energy K and solving for d yields:
d = K / (µk * m * g) = 833.25 J / (0.20 * 150 kg * 9.81 m/s^2) = 833.25 / (294.3) = 2.83 m.