1. An 80kg football player is running at the speed of 6.0m/s. A 100kg player is running towards the other player with a speed of 3.0m/s. At what speed are they travelling when they collide? Give the direction.


2. A reckless driver is driving __m/s and hits the break __m before the red light. He has a deceleration of -16m/s.
a) When does he come to a complete stop in the middle of the intersection?
b) When did he need to brake to make it before the intersection?

3. A driver is driving off a cliff with a speed of 2.0m/s.
a) How far down does he fall? (I think)
b) immediately after driving off the cliff, a rock falls into the water. At what speed does the rock hit the water?

1. Given:

M1 = 80kg, V1 = 6.0m/s.
M2 = 100kg, V2 = -3.0m/s.
V3 = Velocity of M1 and M2 after colliding.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V3.
80*6 + 100*(-3) = 80V3 + 100V3,
480 - 300 = 180V3,
V3 = 1m/s in the direction of M1 before colliding.

2.

1. To find the speed at which the two football players collide, we can use the law of conservation of momentum. The equation is:

(m1 * v1) + (m2 * v2) = (m1+m2) * v

Where:
m1 = mass of player 1 = 80 kg
v1 = velocity of player 1 = 6.0 m/s
m2 = mass of player 2 = 100 kg
v2 = velocity of player 2 = -3.0 m/s (negative because player 2 is running towards player 1)
(v) = final velocity of both players after collision (this is what we need to find)

Substituting the values into the equation, we get:

(80 kg * 6.0 m/s) + (100 kg * (-3.0 m/s)) = (80 kg + 100 kg) * v

(480 kg·m/s) - (300 kg·m/s) = (180 kg) * v

180 kg·m/s = (180 kg) * v

Dividing both sides by 180 kg, we find:

v = 1.0 m/s

Therefore, the two football players will collide at a speed of 1.0 m/s. The direction of the collision will depend on the initial directions of the players, but typically we consider the positive direction as the direction in which player 1 was running.

2. a) To determine when the reckless driver comes to a complete stop, we can use the equation of motion:

v = u + at

Where:
v = final velocity (which will be 0 m/s when the driver comes to a stop)
u = initial velocity of the driver
a = acceleration (in this case, deceleration)
t = time

Rearranging the equation to solve for time (t):

t = (v - u) / a

Substituting the given values:
u = initial velocity = __ m/s (please provide a value)
a = deceleration = -16 m/s
v = 0 m/s (since the driver comes to a stop)

We can then find the time it takes for the driver to come to a complete stop.

b) To determine when the driver needed to brake to make it before the intersection, we need to find the distance traveled by the driver before coming to a stop. We can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance traveled
u = initial velocity
t = time
a = acceleration

We know that the final velocity is 0 m/s and the acceleration is -16 m/s. We need to solve for time (t) and then use that value to find the distance traveled (s).