1.) A 10,000 kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed afterward?

2.) A 3.0 kg ball at 3 m/s and strikes a 1.5 kg ball moving at -2 m/s. After the collision, the 1.5 kg ball moves to the right at 2 m/s. What is the final velocity of 3.0 kg ball?

3.) A 85 kg running back is moving at +12 m/s and strikes a 125 kg linebacker moving at -8.0 m/s. What is the final velocity if they stick together?

Formula: m1vi+m2vi=m1vf+m2vf

:) help me please!!!

if the collision between the baseball and the bat lasted 1.00ms(milliseconds).what force did the bat exert on the baseball

Sure, I can help you with these questions! To solve these problems, we can use the principle of conservation of momentum. According to this principle, the total momentum before a collision is equal to the total momentum after the collision. In other words, the sum of the initial momenta of all objects involved in the collision is equal to the sum of their final momenta.

Let's first look at question 1:

Given:
Mass of first car (m1) = Mass of second car (m2) = 10,000 kg
Initial velocity of first car (v1i) = 24.0 m/s
Initial velocity of second car (v2i) = 0 m/s (at rest)
Final velocity of both cars (vf) = ?

Using the principle of conservation of momentum, we can write the equation:
m1 * v1i + m2 * v2i = (m1 + m2) * vf

Substituting the values into the equation:
(10,000 kg * 24.0 m/s) + (10,000 kg * 0 m/s) = (10,000 kg + 10,000 kg) * vf

240,000 kg*m/s = 20,000 kg * vf

Dividing both sides of the equation by 20,000 kg:
vf = 240,000 kg*m/s / 20,000 kg
vf = 12.0 m/s

Therefore, the common speed of the cars after the collision is 12.0 m/s.

Now let's move on to question 2:

Given:
Mass of first ball (m1) = 3.0 kg
Initial velocity of first ball (v1i) = 3 m/s
Mass of second ball (m2) = 1.5 kg
Initial velocity of second ball (v2i) = -2 m/s
Final velocity of first ball (vf) = ?

Using the same principle of conservation of momentum, we get:
m1 * v1i + m2 * v2i = m1 * vf + m2 * vf

Substituting the values into the equation:
(3.0 kg * 3 m/s) + (1.5 kg * -2 m/s) = 3.0 kg * vf + 1.5 kg * vf

9 kg*m/s - 3 kg*m/s = vf * (3.0 kg + 1.5 kg)
6 kg*m/s = vf * 4.5 kg

Dividing both sides of the equation by 4.5 kg:
vf = 6 kg*m/s / 4.5 kg
vf = 1.33 m/s (rounded to two decimal places)

Therefore, the final velocity of the 3.0 kg ball is approximately 1.33 m/s.

Finally, let's solve question 3:

Given:
Mass of running back (m1) = 85 kg
Initial velocity of running back (v1i) = 12 m/s
Mass of linebacker (m2) = 125 kg
Initial velocity of linebacker (v2i) = -8.0 m/s
Final velocity of both players (vf) = ?

Again, using the principle of conservation of momentum:
m1 * v1i + m2 * v2i = (m1 + m2) * vf

Substituting the values into the equation:
(85 kg * 12 m/s) + (125 kg * -8.0 m/s) = (85 kg + 125 kg) * vf

1,020 kg*m/s - 1,000 kg*m/s = 210 kg * vf

Dividing both sides of the equation by 210 kg:
vf = 20 kg*m/s / 210 kg
vf = 0.095 m/s (rounded to three decimal places)

Therefore, the final velocity of both players together is approximately 0.095 m/s.

I hope this explanation helps! Let me know if you have any further questions.