Conservation of momentum: throwing bricks

In summary, a 59.8kg person standing on a 2.14kg skateboard, holding two 0.902kg bricks, throws each brick with a speed of 16.7m/s relative to the person. The final speed of the person and the skateboard relative to the ground is 0.486 m/s, which is negative because a negative sign was dropped and also because the problem states that the bricks are thrown one at a time, making the speed relative to the ground different for each throw.
  • #1
henry3369
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0

Homework Statement


A 59.8kg person holding two 0.902kg bricks stands on a 2.14kg skateboard. Initially, the skateboard and the person are at rest.

Calculate the final speed of the person and the skateboard relative to the ground if the person throws the bricks one at a time. Assume that each brick is thrown with a speed of 16.7m/s relative to the person.

Homework Equations


m1v1+m2v2 = m1v1'+m2v2'

The Attempt at a Solution


0 = m1v1' + m2v2' + m3v3'
0 = (59.8+2.14)(v1') + (.902)(16.7)+(.902)(16.7)
v1' = 0.486 m/s

Why is this incorrect? The initial momentum is zero because they are at rest, then if I consider the system as the skateboard, bricks, and the person, the final momentum has to equal zero too. So the momentum of the two bricks, each moving at 16.7 m/s, and the momentum of the skateboard and the person should equal zero.
 
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  • #2
One obvious point- v1' will have to be negative. The solution to the equation will have to be negative. You have dropped a negative sign when you solved the equation.

Much more important- you have misread the problem. You have done this as if both bricks were thrown at the same time, when the person is stationary so that "speed relative to the person" is the same as "speed relative to the ground". But the problem says that the bricks are thrown one at a time and both at 26.7 m/s relative to the person. When the first brick is thrown that is the same as "relative to ground" but not when the second brick is thrown.
 

Related to Conservation of momentum: throwing bricks

1. What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant, unless acted upon by an external force. This means that in any interaction between objects, the total momentum before the interaction is equal to the total momentum after the interaction.

2. How does conservation of momentum apply to throwing bricks?

When you throw a brick, you are creating an interaction between the brick and your hand. Before the throw, the total momentum of the system (brick + hand) is zero. When you release the brick, it gains momentum in the direction of the throw, and your hand loses an equal amount of momentum in the opposite direction. This allows for the conservation of momentum to hold true.

3. Is conservation of momentum always true?

Yes, conservation of momentum is considered a universal law in physics and holds true in all interactions between objects. It has been extensively tested and has been shown to hold true even in extreme conditions, such as at the subatomic level or in outer space.

4. What are some real-life applications of conservation of momentum?

Conservation of momentum has many practical applications in everyday life, such as in sports (e.g. throwing a ball or running), transportation (e.g. car accidents), and engineering (e.g. designing rockets). It is also crucial in understanding collisions, explosions, and other types of interactions between objects.

5. Can conservation of momentum be violated?

In theory, conservation of momentum cannot be violated. However, in certain situations, it may appear to be violated due to external forces that are not taken into account. For example, in a car crash, the total momentum of the system may seem to change, but this is because the force of the impact is not the only force acting on the system (e.g. friction, air resistance). In reality, the total momentum of the system remains constant.

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