What is the velocity of the 10.0 kg object after the collision?

In summary, a 15.0 kg object is moving east at 7.0 m/s when it collides with a 10.0 kg stationary object. After collision, the 15.0 kg object is traveling at 4.2 m/s at 340 degrees. The 10.0 kg object is traveling at 5.1 m/s at 25 degrees.
  • #1
scorpa
367
1
Velocity After Collision

Hello everyone!
I am new to these forums, and it looks like they are a great place to learn and ask questions! I have been given an assignment to do and there are a few questions that I can't get, if anyone could help me I would really appreciate their efforts. Oh, and sorry for posting in the wrong forum the first time! :redface:

#1
(A) A 15.0 kg object is moving east at a velocity of 7.0 m/s when it collides with a 10.0 kg stationary object. After collision the 15.0 kg object is traveling at a velocity of 4.2 m/s at 340 degrees. What is the velocity of the 10.0 kg object after the collision?

I listed all of my givens but still can't seem to find an answer. By the way the answer to the question is 5.1 m/s at 25 degrees.

(B) Is this an elastic or inelastic collsion. Provide mathematical evidence for your answer.

I know that this is an inelastic collision because the kinetic energy will change and be lost as heat, sound, and light energy. But how do I prove it mathematically? The answer is 1.1 x 10^2 J . I was going to use the equation Ek= (1/2)m(v^2), but as I can't figure out part A it isn't helping me much.

If anyone can help me I would really appreciate it the effort!
 
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  • #2
Once you figure out part A, you can substitute into that formula to see if kinetic energy was conserved.

For A, split the total initial momentum and the total final momentum into x and y components. The momentum in each direction is conserved. You should be able to get the x and y components of the velocity of the 10.0-kg object after the collision, which can be added vectorally to find the answer.
 
  • #3
scorpa said:
Velocity After Collision

Hello everyone!
I am new to these forums, and it looks like they are a great place to learn and ask questions! I have been given an assignment to do and there are a few questions that I can't get, if anyone could help me I would really appreciate their efforts. Oh, and sorry for posting in the wrong forum the first time! :redface:

#1
(A) A 15.0 kg object is moving east at a velocity of 7.0 m/s when it collides with a 10.0 kg stationary object. After collision the 15.0 kg object is traveling at a velocity of 4.2 m/s at 340 degrees. What is the velocity of the 10.0 kg object after the collision?

I listed all of my givens but still can't seem to find an answer. By the way the answer to the question is 5.1 m/s at 25 degrees.

(B) Is this an elastic or inelastic collsion. Provide mathematical evidence for your answer.

I know that this is an inelastic collision because the kinetic energy will change and be lost as heat, sound, and light energy. But how do I prove it mathematically? The answer is 1.1 x 10^2 J . I was going to use the equation Ek= (1/2)m(v^2), but as I can't figure out part A it isn't helping me much.

If anyone can help me I would really appreciate it the effort!
For (A), use conservation of linear momentum laws. Horizontal momentum is always conserved (initial horiz momentum = final horiz momentum). Same with vertical momentum. So just split the momentum for each object into its x and y components.
For (B), just find the total initial kinetic energy of both objects and compare it to the final total kinetic energy. If they are different, it's an inelastic collision.
 
  • #4
Sorry Siru, didn't mean to repeat what you said. I did'nt see your post... :tongue2:
 
  • #5
This is what I tried to do, I got an answer that was close to the correct one but still now close enough. I will try it again and see what happens. Thank you both for your help, I appreciate it!
 
  • #6
OK I tried it again and this is what happened.

Object A) m=15 kg Vi= 7m/s @ 0 degrees Vf= 4.2 m/s @ 340 degrees

momentum initial = 15kg x 7m/s = 105 kg*m/s
momentum final = 15 x 4.2 = 63

Object B) m= 10kg Vi= 0m/s stationary Vf= ?

momentum initial is at 0 as the object is at rest

x y
105cos0 105
105sin0 0
63cos340 59.2
63sin340 -21.5

total 164.2 -21.5

So now to find the resultant i used pythagoreans theorum and ended up with 165.6

then assuming momentum before equals momentum after I went like this

mv= 165.6
10 x v = 165.6
v= 16.56 m/s which is completely wrong. I don't know what I am doing. I just started physics yesterday and it is taking so long for this stuff to come back to me. If someone could take me through it step by step I would really appreciate the help, I am so lost right now. I could get every other question but his one...arghhh.
 
  • #7
ahhhh I figured it out...turns out I'm quite stupid...haha...I can't believe I never got it in the first place.
 

What is "Velocity After Collision #2"?

"Velocity After Collision #2" refers to the velocity of an object after it has collided with another object. It is a measurement of the speed and direction of the object's movement after the collision has occurred.

How is "Velocity After Collision #2" calculated?

The calculation for "Velocity After Collision #2" involves taking into account the mass, velocity, and direction of both objects involved in the collision. The equation used is m1v1 + m2v2 = (m1 + m2)v, where m1 and v1 are the mass and velocity of the first object, m2 and v2 are the mass and velocity of the second object, and v is the final velocity after the collision.

What factors affect the "Velocity After Collision #2"?

The "Velocity After Collision #2" can be affected by various factors such as the mass and velocity of the objects involved, the angle of collision, and any external forces acting on the objects. The type of collision, whether it is elastic or inelastic, also plays a role in determining the final velocity.

How is "Velocity After Collision #2" important in physics?

"Velocity After Collision #2" is important in physics because it helps in understanding the conservation of momentum and energy. It also allows us to predict the motion of objects after a collision and helps in studying the laws of motion.

Can "Velocity After Collision #2" be measured experimentally?

Yes, "Velocity After Collision #2" can be measured experimentally by conducting collision experiments and recording the before and after velocities of the objects involved. This data can then be used to calculate the final velocity using the above-mentioned equation.

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