Vertical Wall approaching Man - Impulse and Momentum Conservation

[Null_Void]

Homework Statement

See pic below.

Relevant Equations

$V_x = 8 + u$

$V_y = 6$

Let the velocity of the wall be $u$

Since there is no impulse in the vertical direction, the vertical motion of the ball is unaffected by the collision.

I proceeded by working in the frame of the moving wall, therefore the ball now Initially has :
$V_x = 8 + u$
$V_y = 6$

For the ball to reach back directly into the hands of the thrower, it must mean that at the time of collision $V_y = 0$ or that the ball is at its highest postion.

Since the collision is elastic, the $V_x$ of ball is simply reversed.

Now to find the velocity $u$, Since the vertical motion remains unaffected and the ball reaches the thrower, We can derive a relation using the time period of the motion:

Time Period = $2V_y/g$

= $10/V_x + 10/V_x$

Therefore,

$\frac {6} {5} = \frac {20} {8+u}$

$u = 26/3$

But the right answer is $13/3$

I'm skeptical about whether the horizontal speed would be the same after the collision with the wall in the wall's frame, but I can't exactly prove why I'm wrong.

 

[anuttarasammyak]

It seems simpler to me to consider in IFR where the wall is at rest, say at x=0.
Position of the thrower
$$x=-L+\mu t$$
speed of the ball
$$v_x=\mu + V \cos \theta$$
$$v_y= V \sin \theta-gt$$
If there were no wall the trajectory should be ... etc.

Spoiler

The thrower and the ball meet again
$$-L+\mu t_1=-[-L+(\mu + V cos \theta)t_1]$$
where
$$t_1=\frac{2V \sin \theta }{g}$$
which is time the ball has the initial height.

 

[haruspex]

For the ball to reach back directly into the hands of the thrower, it must mean that at the time of collision Vy=0 or that the ball is at its highest position.

That would be true if the horizontal speed were the same, but it won’t be.

 

[Null_Void]

That would be true if the horizontal speed were the same, but it won’t be.

I know this is where I'm going wrong, but why wouldn't the horizontal speed be same?

 

[kuruman]

I know this is where I'm going wrong, but why wouldn't the horizontal speed be same?

When a ball bounces off a massive stationary wall elastically, the component of the velocity perpendicular to the wall reverses direction. When the wall is moving, the ball's horizontal component will have a larger/(smaller) magnitude if the wall is moving towards/(away from) the ball. If the wall is moving away from the ball at a speed greater than or equal to the ball's horizontal speed, there will be no collision and no change in the ball's velocity.

 

[Null_Void]

When a ball bounces off a massive stationary wall elastically, the component of the velocity perpendicular to the wall reverses direction. When the wall is moving, the ball's horizontal component will have a larger/(smaller) magnitude if the wall is moving towards/(away from) the ball. If the wall is moving away from the ball at a speed greater than or equal to the ball's horizontal speed, there will be no collision and no change in the ball's velocity.

I understand what you mean, I think it's similar to the basketball and tennis ball example. But then now that I think about it, since we already keep the wall at rest why can't we simply assume that the magnitude before and after will be the same. Or is it wrong to analyse impulses in different frames of reference?

 

[kuruman]

. . . I think it's similar to the basketball and tennis ball example.

I don't know this example.

. . . why can't we simply assume that the magnitude before and after will be the same.

We can, but what magnitude is this that is the same before and after? Answer: The magnitude of the relative horizontal velocity, which does not change from one inertial frame to another.

 

[Lnewqban]

... But then now that I think about it, since we already keep the wall at rest why can't we simply assume that the magnitude before and after will be the same. Or is it wrong to analyse impulses in different frames of reference?

If you keep the wall at rest, then the hands must be moving toward the wall and the rebounce is going to overshoot the hands.
The key is the angle of impact.

Consider that the vertical wall is unable to throw the ball back choosing the angle, like the person did.
In a perfectly elastic collision, it can only reflect the velocity vector, with the same symmetrical angle with which a mirror reflects a ray of light.

 

[haruspex]

since we already keep the wall at rest

If you are working in the frame of reference of the wall then, yes, the rebound horizontal speed is the same. But now the thrower is moving, so the distance back to the thrower is less, so, again, the ball cannot be striking the wall horizontally.

 

[Steve4Physics]

@Null_Void, I’d like to add this to what’s already been said.

For a perfectly elastic collision:
relative speed of approach = relative speed of separation.

Consider a simple 1D example.

A moving wall has velocity u=1m/s relative to the ground.

Before impact, a ball has velocity v = -8m/s relative to the ground,

Relative speed of approach |v-u| = |-8-1| = 9m/s.
(An observer on the wall sees the ball approaching at a speed of 9m/s.)

Since we have a perfectly elastic collision, the relative speed of separation is also 9m/s.
(An observer on the wall will see the ball departing at a speed of 9m/s.)

This means that after the impact, the ball has velocity relative to the ground of = 9+u = 9+1 = 10m/s. The ball’s final velocity relative to the ground is 2u – v.

Note that if the wall had been stationary (u=0), the ball’s final velocity relative to the ground would have been -v = 8m/s, a simple reversal of velocity.

 

[kuruman]

I'm skeptical about whether the horizontal speed would be the same after the collision with the wall in the wall's frame, but I can't exactly prove why I'm wrong.

Maybe you can't prove why you are wrong, but I think it will be useful to you to derive formally what is right, namely the final velocities of the two elastically colliding masses in terms of the initial velocities. Start with the conservation of momentum and energy equations for 1D collisions
$$\begin{align} & m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f} \\
& \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2=\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2. \\
\end{align}$$ Next solve the system of two equations for the two unknown final velocities $v_{jf}$ in terms of the initial velocities $v_{ji}$, $j=1,2.$ Because of the quadratic equation (2), there are two solutions. One of them is $v_{1f}=v{1i}$, $v_{2f}=v{2i}$. This is the "no collision" solution, i.e. the masses go through each other, which obviously satisfies the conservation equations but of no interest here. You need to find the other solution.

Yes, you can find the answer on the web, however this is a derivation that everyone who studies physics needs to do at least once. After you obtain the appropriate expressions, calculate their limit when one of the masses is much greater than the other one, e.g. $m_2>>m_1.$