# Conservation of momentum, change in velocity due to added mass from above

• bloodlust
In summary, the problem involves a 15 kg cart moving east at 3.0 m/s and 10 kg of newspaper being dropped vertically into the cart. The task is to find the final velocity of the cart. The equation m1v1i+m2v2i=m1v1f+m2v2f and kinematic equations and trigonometry may be used to solve the problem. It is suggested to use the momentum before and after the newspaper is added to the cart to find the final velocity, taking into consideration any other factors that may affect the outcome.

## Homework Statement

A 15 kg cart is moving east at 3.0 m/s when 10 kg of newspaper is vertically dropped into the cart. Find final veocity?2. Homework Equations [/b
m1v1i+m2v2i=m1v1f+m2v2f
Kinematic eqs and trig will be necessary to find components for the 10kgnewspaper

## The Attempt at a Solution

Using the eqs isted above

bloodlust said:

## Homework Statement

A 15 kg cart is moving east at 3.0 m/s when 10 kg of newspaper is vertically dropped into the cart. Find final veocity?

2. Homework Equations [/b
m1v1i+m2v2i=m1v1f+m2v2f
Kinematic eqs and trig will be necessary to find components for the 10kgnewspaper

## The Attempt at a Solution

Using the eqs isted above

So what exactly is your question/difficulty. There is no evidence that you have been "Using the eqs isted above" ??

I just don't know where to start. Do i just add the massafter the newspaper is added and find the velocity then?

Give it a try, momentum before (the cart) = momentum after (cart + newspaper) and see what you get.
Does it sound sensible

, we can apply the conservation of momentum principle to solve for the final velocity of the cart after the newspaper is dropped in.

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 is the mass of the cart (15 kg), v1i is the initial velocity of the cart (3.0 m/s), m2 is the mass of the newspaper (10 kg), and v2i is the initial velocity of the newspaper (0 m/s).

Since the newspaper is dropped vertically into the cart, it will have a downward velocity of 9.8 m/s^2 due to gravity. Using trigonometry, we can find the horizontal component of the newspaper's velocity to be 0 m/s.

Now, we can plug in the values into the conservation of momentum equation:

(15 kg)(3.0 m/s) + (10 kg)(0 m/s) = (15 kg)(v1f) + (10 kg)(9.8 m/s^2)

Solving for v1f, we get a final velocity of 2.0 m/s for the cart after the newspaper is dropped in. This means that the cart will continue moving east, but at a slightly slower speed due to the added mass from the newspaper.

In conclusion, the conservation of momentum principle allows us to calculate the final velocity of the cart after the newspaper is dropped in, taking into account both the mass and initial velocity of the cart and the added mass and velocity of the newspaper.

## 1. What is the conservation of momentum?

The conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system (a system with no external forces acting on it) remains constant over time. This means that the total momentum before an event, such as a collision, is equal to the total momentum after the event.

## 2. How does the addition of mass from above affect the change in velocity?

The addition of mass from above can cause a change in velocity due to the conservation of momentum. When a new mass is added to a system, the total momentum of the system will increase, which can cause a change in the velocity of objects within the system. This change in velocity can be calculated using the equation: Δv = Δm * (2v/m), where Δv is the change in velocity, Δm is the added mass, v is the initial velocity, and m is the total mass of the system.

## 3. Can the conservation of momentum be observed in everyday life?

Yes, the conservation of momentum can be observed in many everyday situations. For example, when playing billiards, the initial velocity and mass of the cue ball will determine the final velocities of the cue ball and the other balls after a collision. Another example is when a car collides with a stationary object, the total momentum of the car and object will remain constant before and after the collision.

## 4. How is the conservation of momentum related to Newton's Third Law of Motion?

The conservation of momentum is closely related to Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. In the case of momentum, when two objects collide, the force exerted on each object is equal and opposite, resulting in a conservation of momentum in the system.

## 5. Are there any exceptions to the conservation of momentum?

In classical mechanics, the conservation of momentum holds true in all situations. However, in the realm of quantum mechanics, there are exceptions to this law, such as in the case of radioactive decay, where particles can decay into smaller particles with different momenta. Additionally, in situations involving extremely high velocities or strong gravitational forces, the conservation of momentum may not hold true due to the effects of relativity.