What velocity does the particle need to get to point B?

Click For Summary
SUMMARY

The discussion focuses on calculating the initial velocity required for a particle to reach point B from point A, considering both x and y components of motion. The participants confirm that the vertical acceleration can be expressed as sin(α) * 9.82 m/s², while the horizontal motion follows the equation x = V0 * t. It is emphasized that numerical values for gravitational acceleration (g) should not be substituted without knowing the other variables, and units must be included when doing so. The correct notation for the angle should be α instead of a to avoid confusion in calculations.

PREREQUISITES
  • Understanding of basic physics concepts, specifically projectile motion.
  • Familiarity with trigonometric functions, particularly sine.
  • Knowledge of kinematic equations for motion in two dimensions.
  • Ability to manipulate algebraic expressions and units in physics.
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion.
  • Learn how to apply trigonometric functions in physics problems.
  • Research the significance of units in physics calculations.
  • Explore advanced topics in projectile motion, including air resistance effects.
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in solving projectile motion problems effectively.

ClearWhey
Messages
19
Reaction score
4
Homework Statement
A particle is at point A in a certain moment. The particle rolls with an angle of inclination a. It’s velocity direction is parallel to the upper edge. What velocity V0 does it need to reach into the hole B?

Sorry for my english
Relevant Equations
Look at the picture to get a better idea of the problem.
It was a long time ago I did these kind of problems so I’m a bit rusty. The only thing I can think of is divide it up to two parts one x and one y.

In y the acceleration is sin(a)*9.82? Then put that in the equation and solve for t.

In x the there is no acceleration so the formula is x=V0*t, I put in t from y and solve for V0?

Haven’t gotten the answer paper yet so I have no idea if I’m on the right path or wrong. That is why I’m asking here if I’m solving this question right, if not how should I do it?

Thanks in advance for the help!
71EDAA2E-1DA0-4245-9624-950298303FA5.jpeg
 
Physics news on Phys.org
Looks like the right approach.
 
  • Like
Likes   Reactions: ClearWhey
You have given the inclination angle as ##a## and according to the figure the "horizontal" distance between A and B also labelled as ##a##. This might lead to confusing algebra.

Perhaps you meant to use ##\alpha## as the angle?
 
  • Like
Likes   Reactions: ClearWhey
gneill said:
You have given the inclination angle as ##a## and according to the figure the "horizontal" distance between A and B also labelled as ##a##. This might lead to confusing algebra.

Perhaps you meant to use ##\alpha## as the angle?

Yes you are correct.

haruspex said:
Looks like the right approach.

What do you think about my solution?
4382B696-6109-4DF2-B5B3-BEE9211F2C16.jpeg
 
ClearWhey said:
What do you think about my solution?
That is right, but there is no point in substituting a numerical value for g when you don't have numbers for the rest of the variables. Just leave it as g.
And if you do substitute a numerical value you must include the units: ##v_0=a\sqrt{\frac{9.82\sin(\alpha)ms^{-2}}{2b}}##
 
  • Like
Likes   Reactions: ClearWhey and etotheipi
haruspex said:
That is right, but there is no point in substituting a numerical value for g when you don't have numbers for the rest of the variables. Just leave it as g.
And if you do substitute a numerical value you must include the units: ##v_0=a\sqrt{\frac{9.82\sin(\alpha)ms^{-2}}{2b}}##

Alright will keep that in mind for the future and thanks for the help!
 
  • Like
Likes   Reactions: jim mcnamara

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
12
Views
1K
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
2K