What is the equation of motion for two cubes colliding in vertical motion?

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SUMMARY

The discussion focuses on the equations of motion for two cubes colliding during vertical motion. The first cube is thrown upwards with an initial velocity (v), followed by the second cube thrown five seconds later with the same velocity. Both cubes collide at a height of 9 meters. The relevant equations of motion include s = ut + ½at² and v² = u² + 2as, which are essential for determining the time and velocity of the cubes at the point of collision.

PREREQUISITES
  • Understanding of kinematic equations, specifically s = ut + ½at² and v² = u² + 2as.
  • Familiarity with vector notation in physics, particularly for motion equations.
  • Basic knowledge of vertical motion and gravitational acceleration.
  • Ability to analyze motion in a one-dimensional axis.
NEXT STEPS
  • Study the derivation and application of kinematic equations in vertical motion scenarios.
  • Learn about vector representation of motion and how to apply it in physics problems.
  • Explore the concept of relative motion in collisions and its implications in physics.
  • Investigate the effects of initial velocity and time delay on the trajectory of moving objects.
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of motion and collision dynamics in a vertical context.

heaven eye
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child play with cubes , he threw first one vertically to up with velocity of (v), after five seconds he threw second cube vertically to up with the same velocity , two cubes collided at 9 meter height what the velocity of two cube? (they are the same!)


actually i don't know but final answer, i need help please

and thank you all again , again , and again :-p .
 
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heaven eye said:
child play with cubes , he threw first one vertically to up with velocity of (v), after five seconds he threw second cube vertically to up with the same velocity , two cubes collided at 9 meter height what the velocity of two cube? (they are the same!)
You want to use [tex]s = ut + \frac{1}{2}at^2[/tex] and [tex]v^2 = u^2 + 2as[/tex] setting s to 9 each time. Then the two equal each other and you can find a time as you know the acceleration and the time in each case.

The Bob (2004 ©)
 
Last edited:
You should write the equation of motion in vector form

[tex]\vec{r}(t)=\vec{r}_{0}+\vec{v}_{0}t+\frac{1}{2}\vec{a}t^{2}[/tex]

and then choose an axis along which the movement takes place,an origin for this axis, then a sense on this axis and then do the projection of the vector equality on this axis...

This is just the beginning,but it's essential.

Daniel.
 

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