Finding radius of a loop using impulse and momentum

In summary: When I get my final equation and plug in numbers I keep getting the wrong answer and I'm not sure where my mistake is. I've emailed my professor to see if the mistake is on his part or on mine.
  • #1
dragonfall122
3
0

Homework Statement


block 1 with mass of 680 g starts from rest at the top of a frictionless track at a height of
H = 4.20 M above the ground. Block 1 collides with block 2 (m = 175 g) at ground level. The blocks stick together and barely complete one turn around a frictionless loop. find the radius of the loop.


Homework Equations



p = mv
mg = mv2/r
Ui + Ki = Uf + Kf
J1 + J2 = 0

The Attempt at a Solution



I found the velocity of block 1 using conservation of evergy,
Ui + Ki = Uf + Kf
Ki = 0 because the object is at rest and and Uf = 0 because the object is at height = 0, so
Ui = Kf
so mgh = 1/2mv^2 so V = 9.07 m/s before the collision.

I used that velocity to find it's momentum,
p=mv so p= (.680 kg)(9.07 m/s)
so p is 6.17 kgm/s immediately before the collision.

I know that at the top of the loop the normal force working on the two objects is zero, and a FBD shows that the sum of the forces working in the y direction are mg = mv2/r

I know that block 2 has no impulse or momentum before the collision because its at rest.
I know that somehow I'm supposed to find the velocity of the two blocks after the collision, but I don't know how.
 
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  • #2
I know that at the top of the loop the normal force working on the two objects is zero, and a FBD shows that the sum of the forces working in the y direction are mg = mv2/r
Yes this is the correct equation to solve for Radius.

I know that block 2 has no impulse or momentum before the collision because its at rest.
I know that somehow I'm supposed to find the velocity of the two blocks after the collision, but I don't know how.
To find the final velocity you need to use conservation of momentum.
So P1i + P2i = P1f + P2f
the left side of the equation should be easy for you to solve..Just plug in your values. the right hand however takes some manipulation. You have
P1f + P2f and this gives us m1V1f + m2V2f
We know that after the collision the two masses will have the same velocity (because they are stuck together), so we can re-write the equation to look like V(m1+m2). This V is the velocity for both masses stuck together after the collision. Now we have
P1f + P2f = V(m1+m2). Then solve for V
Hope this helps
 
  • #3
MillerGenuine said:
Yes this is the correct equation to solve for Radius.


To find the final velocity you need to use conservation of momentum.
So P1i + P2i = P1f + P2f
the left side of the equation should be easy for you to solve..Just plug in your values. the right hand however takes some manipulation. You have
P1f + P2f and this gives us m1V1f + m2V2f
We know that after the collision the two masses will have the same velocity (because they are stuck together), so we can re-write the equation to look like V(m1+m2). This V is the velocity for both masses stuck together after the collision. Now we have
P1f + P2f = V(m1+m2). Then solve for V
Hope this helps

okay, for some reason my brain doesn't want to follow this.
p1i = (.680)(9.07) = 6.17 kgm/s
p2i = 0 because there is no velocity so I would I get
6.17 = V(m1 +m2) so
v= 6.17/ (m1 +m2) ?
 
  • #4
p1i = (.680)(9.07) = 6.17 kgm/s
p2i = 0 because there is no velocity so I would I get
6.17 = V(m1 +m2) so
v= 6.17/ (m1 +m2) ?
Yes it will be your initial momentum divided by the sum of the masses.

okay, for some reason my brain doesn't want to follow this.
Is there something you still don't understand? make sure you understand how this problem is done. Conservation of momentum is an important concept in mechanics..& you should know how to maniupulate the Conservation of Momentum equation very well. So do some practice.
 
  • #5
MillerGenuine said:
Yes it will be your initial momentum divided by the sum of the masses.


Is there something you still don't understand? make sure you understand how this problem is done. Conservation of momentum is an important concept in mechanics..& you should know how to maniupulate the Conservation of Momentum equation very well. So do some practice.

it certainly seems that way. When I get my final equation and plug in numbers I keep getting the wrong answer and I'm not sure where my mistake is. I've emailed my professor to see if the mistake is on his part or on mine.

Thanks so much for you help!
dragonfall122
 

1. What is impulse and momentum?

Impulse and momentum are two related concepts in physics that describe the motion of an object. Impulse is the force applied to an object over a period of time, while momentum is the product of an object's mass and velocity.

2. How can impulse and momentum be used to find the radius of a loop?

By analyzing the forces acting on an object as it moves through a loop, we can use the principles of impulse and momentum to determine the minimum radius required for the object to complete the loop without falling off.

3. What are the key equations for calculating impulse and momentum?

The equation for impulse is given by I = FΔt, where I is impulse, F is force, and Δt is the change in time. The equation for momentum is p = mv, where p is momentum, m is mass, and v is velocity.

4. Can impulse and momentum be used for any type of loop?

Yes, the principles of impulse and momentum can be applied to any type of loop, as long as the forces acting on the object and the object's mass and velocity are known.

5. Are there any assumptions or limitations when using impulse and momentum to find the radius of a loop?

Yes, the calculations assume that the object is moving through the loop without slipping or sliding, and that there are no external forces acting on the object other than the applied force. Additionally, the calculations may not be accurate for complex or non-uniform loops.

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