How Do Conservation Laws Apply to Everyday Physics Problems?

  • Thread starter iTz3D
  • Start date
  • Tags
    Momentum
In summary: Plugging in the values, we get:(25.0 kg)(5.00 m/s) + (75.0 kg)(0 m/s) = (25.0 kg + 75.0 kg)v3Solving for v3, we get a final speed of 1.25 m/s for the sled.5. To find the recoil speed of the cannon, we can use the same equation as before, but this time the mass of the cannon (m1) includes
  • #1
iTz3D
1
0
1. A rifle with a weight of 30 N fires a 5.0 g bullet with a speed of 300 m/s. Find the recoil speed (in m/s) of the rifle.

2. If the rifle in the previous problem was held by a man with a weight of 700 N, what is the recoil speed (in m/s) of the man (with the rifle)?

3. If the bullet in the previous problem sticks in a 1.0 kg block of wood, what is the speed (in m/s) of the block (plus bullet) after the impact?

4. A 25.0 kg child on a sled slides down a hill, reaching a speed of 5.00 m/s. At the bottom of the hill, the child's inattentive, 75.0 kg father is standing still. When the child reaches the father, the father is knocked onto the sled. What is the sled's new speed (in m/s)?

5. A 300 kg cannon sits on a frozen lake. It fires a 6.00 kg cannonball horizontally with a speed of 250 m/s. With what speed (in m/s) does the cannon recoil from the blast?

6. An ice skating couple are playing on the ice. The 75 kg man gets a good start (5.0 m/s) skating toward the 60 kg woman, who is moving in the same direction at 2.0 m/s. After grabbing her hand and fligging her forward, he is still moving at 1 m/s in the same direction. How fast (in m/s) has he managed to get the woman moving?

7. In the previous question, energy is not conserved. What is the change in kinetic energy during the "collision" (in J)?

8. Which of the following best explains what happened when the man and woman in the previous questions "collided"?
a) They had a perfectly elastic collision
b) They had a perfectly inelastic collision
c) The man converted some of his internal energy into kinetic energy
d) The woman experienced an elastic collision but the man experienced an inelastic collision

9. You find yourself stuck in outer space with nothing but your physics book and a space suit. You have a mass of 100 kg in your suit, and your physics book has a mass of 2.50 kg. In order to make it back to your stationary spaceship before running out of air, you must move at 0.200 m/s. How fast (in m/s) do you have to throw your physics book to achieve this speed?

10. In the previous question, which direction should you throw your book?
a) At your ship
b) Away from your ship
c) Perpendicular to your path
d) In the nearest trash can :-)
 
Physics news on Phys.org
  • #2


Hello and thank you for your interesting forum post! I am happy to help you with your physics questions.

1. To find the recoil speed of the rifle, we can use the law of conservation of momentum, which states that the total momentum of a system remains constant. We can set up the equation as follows:

m1v1 = m2v2

Where m1 is the mass of the rifle, v1 is the initial velocity of the rifle, m2 is the mass of the bullet, and v2 is the recoil velocity of the rifle.

Plugging in the values, we get:

(30 N)(0 m/s) = (5.0 g)(300 m/s) + (30 N)v2

Solving for v2, we get a recoil speed of 5 m/s for the rifle.

2. To find the recoil speed of the man with the rifle, we can use the same equation as before, but this time the mass of the rifle (m1) includes the mass of the man as well.

(30 N)(0 m/s) = (5.0 g)(300 m/s) + (700 N)v2

Solving for v2, we get a recoil speed of 0.42 m/s for the man with the rifle.

3. To find the speed of the block of wood (plus bullet) after the impact, we can use the law of conservation of momentum again, but this time the equation is slightly different:

m1v1 = (m2 + m3)v2

Where m1 is the mass of the bullet, v1 is the initial velocity of the bullet, m2 is the mass of the block, m3 is the mass of the bullet, and v2 is the final velocity of the block (with the bullet stuck in it).

Plugging in the values, we get:

(5.0 g)(300 m/s) = (1.0 kg + 5.0 g)v2

Solving for v2, we get a speed of 30 m/s for the block (plus bullet).

4. To find the sled's new speed after the child and father collide, we can use the law of conservation of momentum again, but this time the equation is slightly different:

(m1v1 + m2v2) = (m1 + m2)v3

Where m1 is the mass of the child, v
 

1. What is momentum?

Momentum is a physics concept that describes the quantity of motion possessed by an object. It is calculated by multiplying an object's mass by its velocity.

2. Why is momentum important?

Momentum is important because it helps us understand and predict the motion of objects. It is also a conserved quantity, meaning that in a closed system, the total momentum remains constant.

3. How is momentum different from velocity?

Momentum and velocity are related concepts, but they are not the same thing. Velocity is a vector quantity that describes an object's speed and direction of motion, while momentum is a vector quantity that describes an object's mass and velocity.

4. How is momentum calculated?

Momentum is calculated by multiplying an object's mass (m) by its velocity (v). In equation form, it is expressed as p = m * v. The SI unit for momentum is kilogram-meters per second (kg*m/s).

5. What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum before an event (such as a collision) is equal to the total momentum after the event. This means that momentum is always conserved, even if the objects in the system undergo changes in velocity or direction.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Mechanics
2
Replies
53
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top