2 contradicting approaches for a 1D elastic collision

[baseballfan_ny]

Homework Statement

A ping-pong ball and a bowling ball collide elastically on a frictionless surface. The
magnitude of the initial velocity of the ping-pong ball is vp,0 and the direction of the
velocity is in the positive x-direction. The magnitude of the initial velocity of the
bowling ball is vb,0 and the direction of the velocity is in the negative x-direction. You
may assume that the mass of the bowling ball is much greater than the mass of the
ping-pong ball.

After the collision, what is the component of the velocity of the ping-pong ball in the x
direction? (pay attention to signs). Express your answer in some or all of the following
variables: vp0 and vb0

Relevant Equations

$\vec p_f = \vec p_i$
$(\vec v_{rel})_i = -(\vec v_{rel})_f$

So I've managed to confuse myself on this problem :)

Since the problem says we can assume $m_p << m_b$, I'm assuming that the velocity of the bowling ball will be unchanged, such that $\vec v_{b,i} = \vec v_{b,f} = -v_{b,0} \hat i$

I started out using the energy-momentum principle, $(\vec v_{rel})_i = -(\vec v_{rel})_f$
$\vec v_{b,i} - \vec v_{p,i} = -(\vec v_{b,f} - \vec v_{p,f})$

In the $\hat i$ direction...
$-v_{b,0} - v_{p,0} = -(-v_{b,0} - v_{p,f})$
$-2v_{b,0} - v_{p,0} = v_{p,f}$

Which seems reasonable to me, that the ping pong ball goes in the opposite direction with greater speed. But when I try to solve the problem just by applying ordinary conservation of momentum...
$m_p*v_{p,0} - m_b*v_{b,0} = m_p*v_{p,f} -m_b*v_{b,0}$
$m_p*v_{p,0} = m_p*v_{p,f}$
$v_{p,f} = v_{p,0}$

And the same thing would be implied by the KE energy condition, since the bowling ball velocity is unchanged so the $v_{b,0}$ terms would cancel.
$\frac 1 2 m_p*v_{p,0}^2 + \frac 1 2 m_b*v_{b,0}^2 = \frac 1 2 m_p*v_{f}^2 + \frac 1 2 m_b*v_{b,0}^2$

How come these two approaches are giving me totally different answers? Thanks in advance!

 

[PeroK]

You ought to do the problem without shortcuts and see what happens. You should find that the first method is valid: using conservation of energy and assuming that the bowling ball is unaffected. But, that that second method is invalid: you cannot assume that the momentum of the bowling ball is unaffected.

Compare this problem to dropping a bouncy ball on a hard surface (i.e. an elastic collision with the Earth): KE of the ball is nearly conserved, but momentum is reversed.

 

[etotheipi]

N.B. in this problem the assumption you actually want to make, whether using the momentum or energy approaches, is that $$m_p \ll m_b \implies m_p + m_b \approx m_b$$

 

[baseballfan_ny]

Got it, thanks y'all! The first method worked without the shortcuts and it makes much more sense now. The 2nd method seems to also have worked without the shortcuts but resulted in a ton of complicated algebra that I'm just going to assume works out :)

 

[haruspex]

This question is a useful illustration of the pitfalls of making approximations early in the analysis. The rigorous approach is to write the precise equation, then make the approximation, and keep track of the order of magnitude of the error introduced.

In the present case, e.g. for momentum, the change in momentum of the bowling ball is $m_b\Delta v_b$. As $m_b\rightarrow \infty$, $\Delta v_b\rightarrow 0$, but that does not tell us what happens to $m_b\Delta v_b$.

 

[ThEmptyTree]

How I like to solve this kind of problems:

Suppose we have two particles colliding perfectly elastic, knowing their initial speeds and the fact that the mass of the second particles is much greater than the mass of the first particle.
$$m_1,~\vec{v_{1,i}}$$
$$m_2\gg m_1,~\vec{v_{2,i}}$$
Then
$$\frac{2(m_1\vec{v_{1,i}}+m_2\vec{v_{2,i}})}{m_1+m_2}=2\Big(\frac{m_1}{m_1+m_2}\vec{v_{1,i}}+\frac{m_2}{m_1+m_2}\vec{v_{2,i}}\Big)=2\Big(\big(1+\frac{m_2}{m_1}\big)^{-1}\vec{v_{1,i}}+\big(1+\frac{m_1}{m_2}\big)^{-1}\vec{v_{2,i}}\Big)$$
Notice how I broke everything into simple fractions so we can make the approximations
$$\frac{m_2}{m_1}\to\infty,~\frac{m_1}{m_2}\to0$$
So the final result is
$$2\vec{v_{2,i}}$$
The equation for the final speeds is
$$v_{k,f}=\frac{2(m_1\vec{v_{1,i}}+m_2\vec{v_{2,i}})}{m_1+m_2}-\vec{v_{k,i}},~k=1,2$$
Using the above result
$$v_{k,f}=2\vec{v_{2,i}}-\vec{v_{k,i}}$$
$$v_{1,f}=2\vec{v_{2,i}}-\vec{v_{1,i}}$$
$$v_{2,f}=2\vec{v_{2,i}}-\vec{v_{2,i}}=\vec{v_{2,i}}$$

This explains why the velocity of the object with much greater mass is unchanged.

Final conclusion:

Do not assume $m_2\to\infty$ but instead $\frac{m_2}{m_1}\to\infty,~\frac{m_1}{m_2}\to0$