A 1.2-kilogram basketball travelling at 7.5 meters per second hits the back of a 12-kilogram wagon and bounces off a 3.8 meters per second, sending wagon off in the original direction of travel of the ball. How fast is the wagon going?

M1*V1 + M2*V2 = M1*V3 + M2*V4.

1.2*7.5 + 12*0 = 1.2*(-3.8) + 12*V4.
9 + 0 = -4.56 + 12V4.
12V4 = 13.56.
V4 = 1.13 m/s. = Velocity of the wagon.

To solve this problem, we can use the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Here's the step-by-step solution:

Step 1: Calculate the momentum before the collision.
The momentum of an object is given by the formula: momentum = mass × velocity.
The momentum before the collision is the sum of the momentum of the basketball and the wagon.
The momentum of the basketball before the collision = mass of basketball × velocity of basketball
= 1.2 kg × 7.5 m/s = 9 kg m/s
The momentum of the wagon before the collision = mass of wagon × velocity of wagon
= 12 kg × 0 m/s (since the wagon was initially stationary)
= 0 kg m/s
The total momentum before the collision = momentum of basketball + momentum of wagon
= 9 kg m/s + 0 kg m/s
= 9 kg m/s.

Step 2: Calculate the momentum after the collision.
Since the basketball bounces off the wagon and sends it off in the original direction of travel, the ball transfers its momentum to the wagon.
The momentum of the basketball after the collision = mass of basketball × velocity of basketball (after collision)
= 1.2 kg × (-3.8 m/s) (since the direction of motion is opposite after the collision)
= -4.56 kg m/s (negative sign indicates opposite direction)
The momentum of the wagon after the collision = mass of wagon × velocity of wagon (after collision)
= 12 kg × v (let's assume the velocity of the wagon after the collision is v)
The total momentum after the collision = momentum of basketball after the collision + momentum of wagon after the collision
= -4.56 kg m/s + 12 kg × v
= -4.56 kg m/s + 12v kg m/s.

Step 3: Apply the law of conservation of momentum.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Total momentum before the collision = Total momentum after the collision
9 kg m/s = -4.56 kg m/s + 12v kg m/s.

Step 4: Solve for the velocity of the wagon.
Rearranging the equation, we have:
12v kg m/s = 9 kg m/s + 4.56 kg m/s
12v kg m/s = 13.56 kg m/s
v = 13.56 kg m/s / 12 kg
v ≈ 1.13 m/s.

Therefore, the wagon is moving at approximately 1.13 meters per second after the collision.

To determine the speed of the wagon after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is defined as the product of its mass and velocity. Mathematically, momentum is represented as:

Momentum = mass × velocity

Before the collision, the basketball and the wagon are separate objects moving with their respective velocities:

Basketball's momentum before collision = mass of basketball × velocity of basketball
Wagon's momentum before collision = mass of wagon × velocity of wagon

After the collision, the basketball bounces off and the wagon moves with a new velocity:

Basketball's momentum after collision = mass of basketball × new velocity of basketball
Wagon's momentum after collision = mass of wagon × new velocity of wagon

According to the conservation of momentum principle, we can write the equation as:

Basketball's momentum before collision + Wagon's momentum before collision = Basketball's momentum after collision + Wagon's momentum after collision

Using the given information:

Mass of basketball = 1.2 kg
Velocity of basketball before collision = 7.5 m/s
Mass of wagon = 12 kg
Velocity of wagon before collision = 0 m/s (since it is initially at rest)

Let's substitute the values into the equation:

(1.2 kg × 7.5 m/s) + (12 kg × 0 m/s) = (1.2 kg × new velocity of basketball) + (12 kg × new velocity of wagon)

Simplifying the equation:

9 kg·m/s = 1.2 kg·m/s × new velocity of basketball + 12 kg × new velocity of wagon

We are given the new velocity of the basketball after the collision, which is 3.8 m/s. Substituting this value into the equation:

9 kg·m/s = 1.2 kg·m/s × 3.8 m/s + 12 kg × new velocity of wagon

Now we can solve for the new velocity of the wagon:

9 kg·m/s - 1.2 kg·m/s × 3.8 m/s = 12 kg × new velocity of wagon

Simplifying:

9 kg·m/s - 4.56 kg·m/s = 12 kg × new velocity of wagon

4.44 kg·m/s = 12 kg × new velocity of wagon

Dividing both sides of the equation by 12 kg:

4.44 kg·m/s / 12 kg = new velocity of wagon

0.37 m/s ≈ new velocity of wagon

Therefore, the wagon is moving at approximately 0.37 meters per second after the collision.

htrth