What is the maximum height reached by the other ball after collision?

  • Thread starter Thread starter Rasine
  • Start date Start date
  • Tags Tags
    Balls Collison
Click For Summary
SUMMARY

The discussion focuses on the maximum height reached by a second steel ball after an elastic collision with a first ball, both having a mass of 1.4 kg and suspended from 36 cm strings. The first ball is pulled back to a 36° angle and released, achieving a velocity of 23.89 m/s just before impact. Using conservation of momentum, the final velocity of the second ball is also calculated to be 23.89 m/s, suggesting it will rise to the same height as the first ball. The calculations involve gravitational potential energy and kinetic energy principles.

PREREQUISITES
  • Understanding of elastic collisions in physics
  • Knowledge of gravitational potential energy and kinetic energy equations
  • Familiarity with trigonometric functions and their applications in physics
  • Ability to apply conservation of momentum principles
NEXT STEPS
  • Study the principles of elastic collisions in detail
  • Learn about energy conservation in mechanical systems
  • Explore trigonometric applications in physics problems
  • Investigate the effects of different angles on collision outcomes
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of elastic collisions and energy conservation in mechanical systems.

Rasine
Messages
208
Reaction score
0
Two identical steel balls, each of mass 1.4 kg, are suspended from strings of length 36 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle q = 36° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?

ok so what i am using is pi=pf and ei=ef

first i calcuate the velocity of of th first ball when it is just about to hit the second which is
mgh=1/2mv^2
(1.4)(9.8)36cos36=1/2(1.4)v^2

then v=23.89

so pi=pf of the system so m1v1+m2v2=m1v1f+m2v2f

which is (1.4)(23.89)+0=0+1.4v2f

so v2f=23.89 and if that is true, wouldn't ball 2 go as high as ball 1?
 
Physics news on Phys.org
Rasine said:
first i calcuate the velocity of of th first ball when it is just about to hit the second which is
mgh=1/2mv^2
(1.4)(9.8)36cos36=1/2(1.4)v^2

Reconsider the red step.
 
would it be 36sin56 instead?
 
Rasine said:
would it be 36sin56 instead?

I suggest you draw a diagram on this one.
 
well i want to find the length in the y direction and i see 2 ight triangles i could get y from 36cos36 or take 90-36=56 and do 36sin56
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
12
Views
7K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
21
Views
3K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K