Classic Hit and Stick momentum?

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In summary: Thanks!In summary, the clay balls collide and stick together. The direction and speed of the resulting blob of clay is determined by the direction and speed of the initial balls.
  • #1
NoobeAtPhysics
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Homework Statement



Three balls of clay have masses m1 = 106 g, m2 = 148 g, m3 = 120 g, and speeds v1 = 1.48 m/s, v2 = 1.19 m/s, v3 = 1.85 m/s. They move in the directions: (1) at 45° from the horizontal, (2) horizontally left, and (3) vertically up. They collide simultaneously and stick together. Calculate the speed and the direction (angle) of the resulting blob of clay.

Homework Equations



p=mv

The Attempt at a Solution



Vx = 106 * 1.48 * cos(45) - 148 * 1.19 = -65.189882 m/s

Vy = 106*1.48sin(45) + 120*1.85 = 332.0309 m/s

V = sqrt(332.0309^2 + [-65.189882]^2) (answer?)

angle = arctan (Vy/Vx)


This isn't right is it,
 
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  • #2
NoobeAtPhysics said:

Homework Statement



Three balls of clay have masses m1 = 106 g, m2 = 148 g, m3 = 120 g, and speeds v1 = 1.48 m/s, v2 = 1.19 m/s, v3 = 1.85 m/s. They move in the directions: (1) at 45° from the horizontal, (2) horizontally left, and (3) vertically up. They collide simultaneously and stick together. Calculate the speed and the direction (angle) of the resulting blob of clay.

Homework Equations



p=mv

The Attempt at a Solution



Vx = 106 * 1.48 * cos(45) - 148 * 1.19 = -65.189882 m/s

Vy = 106*1.48sin(45) + 120*1.85 = 332.0309 m/s

V = sqrt(332.0309^2 + [-65.189882]^2) (answer?)

angle = arctan (Vy/Vx)


This isn't right is it,
First of all we have to know the direction of the 45° ball. Is it 45° to the left or to the right?

It is not clear what you are doing in calculating vx and vy.

I suggest you write out the equations for momentum before and after the collision and then relate them. How are they related?

AM
 
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  • #3
No that isn't right, remember that the conservation of momentum applies, use it. Calculate the total momentum in the x direction before the collision, then it them equal to the total momentum in the x direction after the collision, then do the same for the y direction. Then you will have two equations with two unknowns, use algebra to solve for both. What you did was incorrect, you cannot find the total momentum in the x direction and set it equal to the velocity in the x direction. If in doubt, check the units.
 
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  • #4
So

mv = m1 * v1 + m1*v2 + m2*v3? in the x and y directions then use trig to find speed and direction?
 
  • #5
NoobeAtPhysics said:
So

mv = m1 * v1 + m1*v2 + m2*v3? in the x and y directions then use trig to find speed and direction?
Yes. What you calculated in the OP (assuming m1 was moving up and to the right, which is not clear) were the x and y momenta, not the x and y speeds. You just needed to divide by the combined mass.
 
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  • #6
Cool, I got it!
 

Related to Classic Hit and Stick momentum?

1. What is "Classic Hit and Stick momentum"?

"Classic Hit and Stick momentum" is a term used in physics to describe the transfer of momentum between two objects during a collision. It refers to the idea that a stationary object (the "stick") will experience a change in momentum when hit by a moving object (the "hit").

2. How is "Classic Hit and Stick momentum" calculated?

The momentum of an object can be calculated by multiplying its mass by its velocity. In the case of "Classic Hit and Stick momentum," the momentum of the hit object is transferred to the stick object, resulting in a change in the stick's momentum. This change in momentum can be calculated using the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.

3. What factors can affect "Classic Hit and Stick momentum"?

Several factors can affect "Classic Hit and Stick momentum," including the masses and velocities of the objects involved in the collision, the angle of the collision, and the elasticity of the objects. A more massive or faster-moving object will have a greater momentum and therefore a greater impact on the stick. The angle of the collision can also affect the transfer of momentum, with a head-on collision resulting in a larger change in momentum compared to a glancing blow. The elasticity of the objects can also impact the transfer of momentum, with more elastic objects bouncing off each other and transferring more momentum compared to less elastic objects that stick together after the collision.

4. What is the significance of "Classic Hit and Stick momentum" in physics?

Understanding "Classic Hit and Stick momentum" is important in physics because it helps explain how objects interact with each other during collisions. It is a crucial concept in fields such as mechanics and engineering, where the transfer of momentum is essential in designing structures or predicting the outcome of collisions. Additionally, the law of conservation of momentum, which is demonstrated through "Classic Hit and Stick momentum," is a fundamental principle in physics and has many applications in various scientific fields.

5. Can "Classic Hit and Stick momentum" be observed in real-life situations?

Yes, "Classic Hit and Stick momentum" can be observed in many real-life situations, such as in sports, car accidents, or even in everyday activities like playing billiards. In sports like football or hockey, players use "Classic Hit and Stick momentum" to transfer momentum to the ball or puck, increasing its speed and distance. In car accidents, the transfer of momentum between two vehicles can determine the severity of the collision. Understanding "Classic Hit and Stick momentum" can also help us make predictions and prevent accidents in these scenarios.

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