Conservation of momentum (with conservation of energy)

Click For Summary
SUMMARY

The discussion centers on a physics problem involving a 17.00 kg sphere and a 6.00 kg dart, where the dart embeds itself in the sphere after a horizontal strike. The objective is to determine the minimum initial speed of the dart required for the combined mass to complete a circular loop after the collision. The calculations reveal that the velocity needed at the top of the loop is 5.943 m/s, leading to a total energy of 2436.87 Joules at that point. However, the user incorrectly calculates the dart's required speed as 55.8 m/s, indicating a misunderstanding in applying the conservation of momentum principle.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with centripetal force and gravitational force equations
  • Knowledge of kinetic and potential energy calculations
  • Ability to solve equations involving square roots and algebraic manipulation
NEXT STEPS
  • Review the principles of conservation of momentum in inelastic collisions
  • Study the relationship between centripetal force and gravitational force in circular motion
  • Learn how to calculate total mechanical energy in a system
  • Practice solving similar physics problems involving energy and momentum
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of conservation laws in action.

S[e^x]=f(u)^n
Messages
23
Reaction score
0

Homework Statement


A 17.00kg sphere is hanging from a hook by a thin wire 3.60m long, it is free to swing in a complete circle. Suddenly it is struck horizontally by a 6.00kg dart that embeds itself in the sphere. What is the minimum initial speed of the dart so the combination makes a complete circular loop after the collision?

Homework Equations


F_c\mbox{top of loop} = T+m_c g\\<br /> E_k=frac{mv^2}{2}\\<br /> E_g=mg\Delta h[\tex]<h2>The Attempt at a Solution</h2><br /> So basically what i did was figured out the velocity at the top of the loop needed to produce 0 tension since the object must make a full loop and not just get to the top.<br /> <br /> i did this by setting the centripetal force equal to the force of gravity and the tension<br /> since we know tension is equal to 0, the centripetal force is then equal to the gravitational force acting on the object. the masses cancel and this gives...<br /> frac{mv^2}{r}=mg\\v=sqrt{gr}[\tex] &lt;br /&gt; &lt;br /&gt; we know both r and g so we can then calculate the velocity at the top of the circle needed for 0 tension in the line to be 5.943m/s. Using this, we can then calculate the total energy of the particle at that point (knowing the kinetic and gravitational potential energy)&lt;br /&gt; E_total = frac{mv^2}{2} + m g \delta h[\tex] (energy at the top of the loop)&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; this total energy if my calculations are right is equal to 2436.87 Joules. Since there&amp;amp;#039;s no way the system could have lost energy over the swing, i presumed that at the bottom the energy was the same(without the gravitational potential energy)... this means that the total energy must be equal to only the kinetic energy of the combined particle and dart at the bottom. so you set kinetic energy equal to 2436.87Joules and solve for the velocity of the particle at the bottom.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; i know know the speed at which the particle must be traveling at the bottom. so in order to find the speed at which the dart must hit that particle (and then embed itself in) becomes a simple conservation of momentum question. where the mass of the dart and its velocity (the unknown) is the only momentum in the system prior to impact, and the only momentum after impact is the combined mass of the dart and sphere multiplied by the initial velocity needed to overcome the loop. i get an answer of 55.8m/s for the dart... but its wrong. where and how did i go wrong? sorry for the length of this post.
 
Last edited:
Physics news on Phys.org
bump :(... anybody?
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
Replies
2
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
21
Views
3K
  • · Replies 4 ·
Replies
4
Views
914
Replies
10
Views
3K
Replies
8
Views
4K