Conservation of Momentum initial velocity curved path

Click For Summary
SUMMARY

The discussion centers on a physics problem involving the conservation of momentum and the dynamics of a block and putty system on a frictionless track. The initial velocity (v0) of a 1 Kg lump of putty is calculated to be 70 m/s, based on the momentum equation m1v0 = (m1 + m2)(vf) and the normal force acting on the block/putty combination at point B, which is 98 N. The radius of the quarter-circle track is specified as R = 5 m, and the analysis includes the application of centripetal acceleration and energy conservation principles.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Knowledge of centripetal acceleration and forces
  • Familiarity with energy conservation equations in physics
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of the conservation of momentum equation in collisions
  • Learn about centripetal force and its applications in curved motion
  • Explore energy conservation in mechanical systems
  • Practice solving problems involving normal forces and motion on curved paths
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of momentum and energy conservation in real-world applications.

pconn5
Messages
6
Reaction score
0

Homework Statement


The frictionless track is made of a straight horizontal section and a quarter-circle with radius R = 5 m. A 1 Kg lump of putty is thrown toward the stationary 4 Kg block, and the block starts to slide with the putty stuck on it after the collision. If the normal force acting on the block/putty from the track is 98 N at point B, find the initial speed v0 of the putty.

Homework Equations


m1v0=(m1+m2)(vf)
...energy equation?

The Attempt at a Solution


I have been trying to use the energy equation to solve this problem but I am getting nowhere. The answer is 70 m/s. I can't seem to figure out what the velocity would be at the bottom of the arc. According to a(norm)=v^2/R, I found that v=9.8995 but don't know what to do with that...
 

Attachments

  • Diagram.JPG
    Diagram.JPG
    7.3 KB · Views: 466
Physics news on Phys.org
Well depending on where B is the velocity would be different.

But say B is just at the bottom of the arc, then on the arc the forces acting are the Normal force R and the weight W. The resultant of these two gives mv^2/r

\frac{mv^2}{r}=R-W

so find v from that.

Then put that into the momentum equation.
 
Ok I got it now. Thanks for the help.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
872
  • · Replies 8 ·
Replies
8
Views
3K
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K