Collision of cballs of clay help

In summary, the conversation includes a series of problems involving collisions and momentum. The first problem involves a collision between two clay balls, while the second problem involves a skateboarder jumping off a skateboard. The third problem is similar to the first, but with different angles involved. The fourth problem is more complex, involving a spaceship breaking into three pieces due to a failed reactor. The conversation also touches on the concept of inelastic collisions and the difference between speed and velocity.
  • #1
oldunion
182
0
All of these problems are the last of a series of ten, which i cannot solve. I've given extensive efforts but am now nearly out of time (12pm est). if anone could be of help please feel free.


A 20 g ball of clay traveling east at 2m/s collides with a 30 g ball of clay traveling 30 degrees south of west at 1m/s. what is the speed of the resulting blob, m/s?

Dan is gliding on his skateboard at 2.00m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 7.00m/s(as measured by an observer on the ground). Dan's mass is 60.0 kg and the skateboard's mass is 6.00kg. how fast is dan going as his feet hit the ground

A 10.0g ball of clay traveling east at 7.00m/s collides and sticks together with a 10.0g ball of clay traveling north at 5.50m/s. what is the speed of the resulting clay

A spaceship of mass 2.3×10^6kg is cruising at a speed of 6×10^6m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.8×10^5kg, is blown straight backward with a speed of 2.3×10^6m/s. A second piece, with mass 7.6×10^5kg, continues forward at 1.3×10^6m/s.
 
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  • #2
Do you agree that 1&3 is basically the same problem?
What do you know of inelastic collisions?
 
  • #3
i just solved the spaceship problem.

they are all momentum problems, and i feel as though i understand them all but something isn't quite right. and i know of ineleastic collisions.

for the skateboard question, i did 60kg(Vf)=6kg*2m/s+60kg*2m/s

for blob of clay question (without degrees) i did (.01kg+.03kg)Vf=.02kg*3m/s+.03kg*2m/s
 
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  • #4
The "blob of clay" questions need you to regard velocities as VECTORS, not scalars as you have done
 
  • #5
could you be more specific, do i find x and y components and then add? I am almost positive I am doing it correctly. for the first clay question... (.01kg +.01kg)Vf=.07kgm/s+.055kgm/s final answer of 6.25 is incorrect
 
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  • #6
Right, in so far as you mean to say you add componentwise.
For example, the first one:
Let the unit vector "east" be [tex]\vec{i}[/tex] and north [tex]\vec{j}[/tex]
Then, the unit vector which makes 30 degrees angle to the unit "west" vector in the southern direction is: [tex]\vec{t}=-\cos(30)\vec{i}-\sin(30)\vec{j}[/tex]
Hence, the final velocity, [tex]\vec{v}_{f}[/tex] fulfills:
[tex](m_{1}+m_{2})\vec{v}_{f}=m_{1}v_{1}\vec{i}+m_{2}v_{2}(-\cos(30)\vec{i}-\sin(30)\vec{j})[/tex]
[tex]m_{1}=20g, v_{1}=2\frac{m}{s}, m_{2}=30g, v_{2}=1\frac{m}{s}[/tex]
 
  • #7
thank you so much, but what about the skateboarder, i can't figure it out.

edit: that equation does not work. but I am out of time. thanks for the help
 
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  • #8
Do you know what speed is?
It is not the same as velocity, I thought you knew the difference.
 
  • #9
their desired answer is in the form of m/s, the correct answer is .410. how did they come to this?
 
  • #10
Well, you must calculate the speed from the velocity and plug in the numbers.
To continue with the one I started with, the final speed, [tex]||\vec{v}_{f}||[/tex]
satisfies:
[tex]||\vec{v}_{f}||=\frac{1}{m_{1}+m_{2}}\sqrt{(m_{1}v_{1}-m_{2}v_{2}\cos(30))^{2}+m_{2}^{2}v_{2}^{2}\sin^{2}(30)}=\frac{1}{20+30}\sqrt{(20*2-30*1*\frac{\sqrt{3}}{2})^{2}+30^{2}*1^{2}*\frac{1}{4}}[/tex]
 

1. What is the purpose of studying the collision of clay balls?

Studying the collision of clay balls can help us understand the principles of physics and how objects interact with each other. It also has practical applications in fields such as engineering and materials science.

2. How do the properties of the clay balls affect the collision?

The properties of the clay balls, such as their mass, velocity, and elasticity, can affect the outcome of the collision. These properties determine the amount of force and energy involved in the collision, which can lead to different results.

3. Can the collision of clay balls be used to model real-life collisions?

Yes, the collision of clay balls can be used as a simplified model to study and understand real-life collisions. By controlling and manipulating the properties of the clay balls, we can simulate different scenarios and gain insights into the physics behind collisions.

4. How does the angle of collision impact the outcome?

The angle of collision can greatly affect the outcome of the collision. If the clay balls collide head-on, the force and energy involved will be different compared to a collision at an angle. This can result in different trajectories and velocities of the clay balls after the collision.

5. What other factors can influence the collision of clay balls?

Other factors that can influence the collision of clay balls include air resistance, surface friction, and the presence of other objects. These factors can affect the motion and behavior of the clay balls during the collision, and should be taken into consideration when conducting experiments.

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