Two blocks sliding and the collision

In summary, the smaller block slides back up along the track after collision reach highest point C. If angle locates point C is 60 degrees, find max distance "s" block 2 travel on the incline.
  • #1
spdowind
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0

Homework Statement



Two blocks are on a frictionless track with a vertical, quarter circle, and 30 degree incline. Released from rest at point A, 1 kg smaller block 1 slides to B and makes an inelastic collision with 5 kg stationary larger block 2. The collision sents block 2 up the ramp. The smaller block slides back up along the track after collision reach highest point C. If angle locates point C is 60 degrees, find max distance "s" block 2 travel on the incline.
[PLAIN]http://img704.imageshack.us/img704/752/picturefv.jpg

Homework Equations


Energy equation : K1+ Ug1+ UE1+Work= K2+Ug2+UE2
Inelastic collision equation: m1v1+ m2v2= m1v1' + m2v2'


The Attempt at a Solution



1. I found the velocity for slider 1 BEFORE collision using energy equation.
2. I then use the inelastic collision equation and found V'1(at B) + 5V'2(at B) = 7.668. Which is the equation for velocity of slider 1 and 2 at point B after they collide.
3. Then I used energy equation again for B to C motion by slider 1 AFTER collision. I found V'1( at B) = 14.7. I plug this in the step 2 result and found V'2 (at B), which is the velocity of slider 2 at Point B AFTER collision.
4. Then I used energy equation for motion B to D for slider 2 AFTER collision. D being furthest part on the incline slider 2 travels on. Here I find "s" being 0.202 meters.
However, the answer is 0.54 meters, I am not sure what I am doing wrong, I doubled checked my steps and my calculations, too.

Hopefully someone could help, thanks!
 
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  • #2
I tried the question and I get 0.54m as well. Exactly actually. I think your mistake is here "V'1( at B) = 14.7". I get sqrt(1.5g) = 3.834. Did you forget to square root it here? Also, don't forget V1 should be plugged in as negative since it's moving to the left.
 
  • #3
jdc15 said:
I tried the question and I get 0.54m as well. Exactly actually. I think your mistake is here "V'1( at B) = 14.7". I get sqrt(1.5g) = 3.834. Did you forget to square root it here? Also, don't forget V1 should be plugged in as negative since it's moving to the left.

I tried it again with cleaner procedures and found that it should be sqrt 14.7. However, I forgot to make V1 negative this time. Good thing I checked this and with your help I changed V1 to negative and guess what, I finally found the correct answer.
Thank you so much! This was the only one I'm stuck on for my test tomorrow and now I am all set!
 

Related to Two blocks sliding and the collision

1. How does the mass of each block affect the collision?

The mass of each block plays a significant role in the collision. The heavier block will have a greater momentum and will exert a larger force on the lighter block, causing it to move faster.

2. What is the role of friction in the collision between two blocks?

Friction can affect the collision in a few ways. It can slow down the movement of the blocks, decrease the force of impact, and even change the direction of the blocks after collision.

3. Does the velocity of the blocks before collision affect the outcome?

Yes, the velocity of the blocks before collision will determine their momentum and the force of impact during the collision. The faster the blocks are moving, the more energy will be transferred during the collision.

4. How is the conservation of momentum and energy applied in a collision between two blocks?

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. This means that the sum of the momentum of both blocks before collision will be equal to the sum of their momentum after collision. Similarly, the law of conservation of energy states that the total energy before collision is equal to the total energy after collision. Therefore, the kinetic energy of the blocks before and after collision will be the same.

5. Can the elasticity of the blocks affect the collision outcome?

Yes, the elasticity of the blocks can affect the collision outcome. In an elastic collision, the blocks will bounce off each other and retain most of their kinetic energy. In an inelastic collision, the blocks will stick together and some of the kinetic energy will be lost as heat and sound. The elasticity of the blocks can also affect the direction of their movement after collision.

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