Two balls one thrown up and other down

[rudransh verma]

Homework Statement

From the building two balls A and B are thrown such that A is thrown upwards and B downwards in vertical direction with same speed. If v_A and v_B are their respective velocities on reaching ground, then
1)v_B>v_A
2) v_A=v_B
3) v_A>v_B
4) velocity depend on mass

Relevant Equations

$E=K+U=constant$

I think because both are launched with same speed so both have same KE. Since one is Thrown downwards it’s KE will increase but not that much as the body which is thrown upwards. Because it covers more distance so it gains more energy(the body thrown up). So 3rd option must be right.

 

[Steve4Physics]

I think because both are launched with same speed so both have same KE.

But their masses could be different. The question doesn't say equal masses.

Since one is Thrown downwards it’s KE will increase but not that much as the body which is thrown upwards.

Assume negligible air resistance. Call the (identical) initial speeds 'V'.

Ball A rises to max. height and then comes down. What will ball A's speed be when (while falling downwards) it is level with its intial launch point? Why?

 

[rudransh verma]

ball A's speed be when (while falling downwards) it is level with its intial launch point? Why?

V. (Conservation of mechanical energy )

 

[Frabjous]

This is essentially identical to your tomato problem from yesterday. You are too focused on pursuing tangents. You need to start working with the equations, rather than trying to talk your way through the problem. Working with equations is how you learn to talk your way through the problem.

 

[Orodruin]

V. (Conservation of mechanical energy )

So it has the same speed downwards when passing the launch point as the other ball had when being thrown from the launch point ... hence ...

 

[phinds]

So 3rd option must be right.

V. (Conservation of mechanical energy )

How is it possible that you do not see that these two statements are mutually contradictory ?

 

[rudransh verma]

This is essentially identical to your tomato problem from yesterday. You are too focused on pursuing tangents. You need to start working with the equations, rather than trying to talk your way through the problem. Working with equations is how you learn to talk your way through the problem.

Ok! On arriving at the launch point body A will have
KE=1/2mv_0^2(initial)= mgh(max height) =1/2mv^2(final).
Then $W=\Delta KE=-mgd=KEfinal-1/2mv_0^2$
For body B same final kinetic energy. So same speed at the ground. Option B is right.

 

[rudransh verma]

But their masses could be different. The question doesn't say equal masses.

I have a confusion. You are right. Then how am I correct?
Final KE must be different because of different masses.

 

[Frabjous]

What does

Ok! On arriving at the launch point body A will have
KE=1/2mv_0^2(initial)= mgh(max height) =1/2mv^2(final).
Then $W=\Delta KE=-mgd=KEfinal-1/2mv_0^2$
For body B same final kinetic energy. So same speed at the ground. Option B is right.

What does work (line 3) have to do with this problem?
Why are you talking about KE (line 4)?
You are writing down equations and guessing a solution.
Solve the equations!

 

[rudransh verma]

What does work (line 3) have to do with this problem?

Work by gravitational force. Same amount of work transfer same energy.

Why are you talking about KE (line 4)?

I thought same energy means same velocity but I am wrong. Now I can see.
If both bodies have same mass then it’s pretty easy to see that delta KE will be same. But here in question it’s not given that.
It’s clear that at the launch point the speed will be same. v^2=u^2-2gh. By third eqn of motion. They will have same velocity/speed. I got it!

 

[Steve4Physics]

I have a confusion. You are right. Then how am I correct?
Final KE must be different because of different masses.

If you drop a ball of mass M and another ball of mass 2M from a tower in Italy (other countries are available) which ball reaches the ground first (assume air resistance is negligible)?

The full answer is in the maths. Can you derive an equation for the final speed of a mass $m$, thrown (in any direction (!)) with initial speed $v_0$ from height $h$?

Once you have this equation, you can see - by inspecting the equation -how changing the mass (or the height or the initial speed) affects the final answer.

 

[fresh_42]

Maybe it helps if you consider a throw which is not perfectly vertical, but at some angle instead.

One ball describes the black plus the red curve, and the other only the red one.

Now, there are three question:

Are the velocities different at the point where black turns into red?
Why does it not depend on the angle?
Why does it not depend on the masses?

 

[rudransh verma]

If you drop a ball of mass M and another ball of mass 2M from a tower in Italy (other countries are available) which ball reaches the ground first (assume air resistance is negligible)?

F=GmM/r^2. a=GM/r^2=g. Same acceleration means same speed near the ground.
v^2=u_0^2-2gh. So same speed at ground. I got it. v does not depend on mass.Thanks.

 

[Steve4Physics]

F=GmM/r^2. a=GM/r^2=g. Same acceleration means same speed near the ground.
v^2=u_0^2-2gh. So same speed at ground. I got it. v does not depend on mass.Thanks.

Sounds like you may have got it. But to complete the process, can you answer the question asked in Post #11, which was:

A mass $m$ is thrown (in an unknown direction) at speed $v_0$ from a height $h$. Derive an equation for the speed of impact of the mass with the ground.

The first line of your derivation could be:
initial total energy = final total energy

 

[rudransh verma]

A mass m is thrown (in an unknown direction) at speed v0 from a height h. Derive an equation for the speed of impact of the mass with the ground.

${v}^2={v_0}^2+2gh$

 

[Steve4Physics]

${v}^2={v_0}^2+2gh$

Yes. Well done. Note:

1) You haven't shown the actual derivation/reasoning.

2) You haven't actually completed the exercise, which requires a formula for $v$, not for $v^2$.

3) From the equation you can tell that changing the mass doesn't affect $v$ (beause mass is not in the formula for $v$).

The exercise was about final speed, so the sign of $v$ is is positive. But, for information, note that your equation gives $v=±\sqrt{{v_0}^2+2gh}$. To find the final velocity for a 1D vertical motion problem, you would need to choose between the positive and negative solutions.

 

[rudransh verma]

Yes. Well done.

Thank you @Steve4Physics I like to be correct. (-ve)