Initial velocity of ball B hit a free falling ball A

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SUMMARY

The problem involves calculating the minimum initial velocity of ball B, which is thrown upward to collide with ball A that is released from a height of 5 meters. Given the gravitational acceleration (g) of 10 m/s², the distance between the balls at release is 4 meters. The correct approach involves using the equation of motion x(t) = v0t + 1/2at² to establish the relationship between the distances traveled by both balls. The minimum initial velocity of ball B is determined to be approximately 4.4 m/s, aligning with the choice of 5 m/s as the closest answer.

PREREQUISITES
  • Understanding of kinematic equations, specifically x(t) = v0t + 1/2at²
  • Basic knowledge of gravitational acceleration (g = 10 m/s²)
  • Concept of relative motion and displacement in physics
  • Ability to solve quadratic equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion.
  • Explore the concept of energy conservation in mechanical systems.
  • Learn about the significance of maximum time in motion problems.
  • Investigate other collision scenarios in physics, such as elastic and inelastic collisions.
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and motion analysis, as well as educators looking for practical examples of collision problems.

Bond
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Homework Statement


A boy release ball A from height 5m from ground at the same time other boy throw ball B vertically upward to exactly strike ball A. When ball B released it was 1m from ground find minimum initial velocity of ball b. Given g=10m/s^2
a) 2m/s b) 3m/s c) 4m/s d) 5m/s

Homework Equations


x (t) = v0t+1/2at^2[/B]

The Attempt at a Solution


From the problem distance between each ball is 4m when it was released and both was released at same time so when it strike each other time must be equal
Then i let x be the distance ball A travel and 4-x be distance ball B travel so i got.
x=1/2gt^2 = 5t^2 since ball A was released it has no initial velocity
and 4-x=v0t-1/2gt^2=v0t-5t^2
Solve for both I got ut =4 then I stuck here then I substitude value from choice and got 5 (the exact is about 4.4) so is there any real way to solve this
P.S. I try using energy conservation and end up with 0=0 all time.[/B]
Sorry if the problem look weird since I translate it from my own language
 
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Hello Bond, :welcome:

Bond said:
since I translate it from my own language
yes, your exercise text puzzles me: what if boy B just holds te ball ? Then ball A will also strike ball B exactly ?

Or does it say that the balls hit each other exactly half way ?
 
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BvU said:
Hello Bond, :welcome:

yes, your exercise text puzzles me: what if boy B just holds te ball ? Then ball A will also strike ball B exactly ?

Or does it say that the balls hit each other exactly half way ?
BvU said:
Hello Bond, :welcome:

yes, your exercise text puzzles me: what if boy B just holds te ball ? Then ball A will also strike ball B exactly ?

Or does it say that the balls hit each other exactly half way ?
It doesn't say where they hit but I'm very sure that the ball B has to be thrown and hit ball A mid air because they were released at the same time
 
BvU said:
Hello Bond, :welcome:

yes, your exercise text puzzles me: what if boy B just holds te ball ? Then ball A will also strike ball B exactly ?

Or does it say that the balls hit each other exactly half way ?

I reckon they have to collide before ball B hits the ground. That's where the minimum velocity comes in.
 
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Bond said:

Homework Statement


A boy release ball A from height 5m from ground at the same time other boy throw ball B vertically upward to exactly strike ball A. When ball B released it was 1m from ground find minimum initial velocity of ball b. Given g=10m/s^2
a) 2m/s b) 3m/s c) 4m/s d) 5m/s

Homework Equations


x (t) = v0t+1/2at^2[/B]

The Attempt at a Solution


From the problem distance between each ball is 4m when it was released and both was released at same time so when it strike each other time must be equal
Then i let x be the distance ball A travel and 4-x be distance ball B travel so i got.
x=1/2gt^2 = 5t^2 since ball A was released it has no initial velocity
and 4-x=v0t-1/2gt^2=v0t-5t^2
Solve for both I got ut =4 then I stuck here then I substitude value from choice and got 5 (the exact is about 4.4) so is there any real way to solve this
P.S. I try using energy conservation and end up with 0=0 all time.[/B]
Sorry if the problem look weird since I translate it from my own language

Some of your equations have ##g## in them and some don't. Is that a typo?

You need to think about displacements rather than distances here. Also, I would set the ground as ##x = 0## and use that as my reference frame.

That said, ##ut = 4m## is correct. Now, you need an idea about how to calculate the minimum ##u## from that. Hint: if ##u## is a minimum, what is ##t##?
 
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PeroK said:
Some of your equations have ##g## in them and some don't. Is that a typo?

You need to think about displacements rather than distances here. Also, I would set the ground as ##x = 0## and use that as my reference frame.

That said, ##ut = 4m## is correct. Now, you need an idea about how to calculate the minimum ##u## from that. Hint: if ##u## is a minimum, what is ##t##?

So it mean that t should be maximum which is the time ball A take to hit the ground is that correct?
 
Bond said:
So it mean that t should be maximum which is the time ball A take to hit the ground is that correct?

That's right.
 
PeroK said:
That's right.
Thanks a lot
 

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