A golf ball traveling 3 m/s to the right has a head-on collision with a stationary bowling ball in a friction-free environment. If the collision is almost perfectly elastic the speed of the golf ball immediately after the collision is:

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- Aug 16th 2006, 10:49 AMLLELLAA golf ball traveling 3 m/s
A golf ball traveling 3 m/s to the right has a head-on collision with a stationary bowling ball in a friction-free environment. If the collision is almost perfectly elastic the speed of the golf ball immediately after the collision is:

- Aug 16th 2006, 11:42 AMtopsquarkQuote:

Originally Posted by**LLELLA**

If the collision is elastic then we are conserving the kinetic energy of the overall system. Since there are no net external forces acting on the system the momentum of the system is also conserved.

So:

$\displaystyle \frac{1}{2}mv_0^2 + \frac{1}{2}MV_0^2 = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$ (KE equation)

$\displaystyle mv_0 + MV_0 = mv + MV$ (Momentum equation)

Where m and v represent the golf ball, M and V represent the bowling ball, and the "0" subscripted variables represent the initial values.

First note that the momentum equation is a*vector*equation so we need to assign a positive direction. I am picturing the golf ball initially moving toward the right at the stationary golf ball. Accordingly I am going to say that the positive direction is to the right. We also know that the bowling ball is initially stationary, thus V0 is 0 m/s.

(Warning! This is going to get a little bit hairy without numbers for the masses!)

$\displaystyle \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$

$\displaystyle mv_0 = mv + MV$

We want to solve this system for v. Solving the second equation for V gives:

$\displaystyle V = \frac{m(v_0 - v)}{M}$

Inserting this value for V into the first equation:

$\displaystyle \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + \frac{1}{2}M \left ( \frac{m(v_0 - v)}{M} \right ) ^2$

$\displaystyle \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + \frac{m^2}{2M} (v_0^2 - 2v_0v + v^2)$

(Reminder, we are solving for v.)

Multiplying both sides by 2M we get:

$\displaystyle mMv_0^2 = mMv^2 + m^2v_0^2 - 2m^2v_0v + m^2v^2$

Rearranging:

$\displaystyle 0 = (m^2 + mM)v^2 + (-2mv_0)v + (m^2v_0^2 - mMv_0^2)$

We can factor a common m from all terms so:

$\displaystyle 0 = (m + M)v^2 + (-2mv_0)v + (m - M)v_0^2$

This is "simply" a quadratic in the variable v, so using the quadratic formula:

$\displaystyle v = \frac{2mv_0 \pm \sqrt{(-2mv_0)^2 - 4(m + M)(m - M)v_0^2}}{2(m + M)}$

The nice thing is that this simplifies a bit:

$\displaystyle v = \frac{2mv_0 \pm \sqrt{4m^2v_0^2 + (4M^2 - 4m^2)v_0^2}}{2(m + M)}$

$\displaystyle v = \frac{2mv_0 \pm \sqrt{4v_0^2 (m^2 + M^2 - m^2)}}{2(m + M)}$

$\displaystyle v = \frac{2mv_0 \pm \sqrt{4M^2v_0^2}}{2(m + M)}$

$\displaystyle v = \frac{2mv_0 \pm 2Mv_0}{2(m + M)}$

$\displaystyle v = \frac{2(mv_0 \pm Mv_0)}{2(m + M)}$

$\displaystyle v = \frac{mv_0 \pm Mv_0}{m + M}$

$\displaystyle v = \frac{m \pm M}{m + M}v_0$

So which sign do we choose? If we choose "+" then:

$\displaystyle v = \frac{m + M}{m + M}v_0 = v_0$

This says that the velocity of the golf ball did not change during the collision. This is a ridiculous assertion. Thus we need to take the "-" sign:

$\displaystyle v = \left ( \frac{m - M}{m + M} \right ) v_0$

Without the masses we can go no further.

NOTE: We can*estimate*the solution if no masses are given. It is reasonable to assume that the mass of the bowling ball is significantly greater than that of the golf ball. In this instance we may neglect the mass of the golf ball in this formula. (Mathematically we are taking the limit where m goes to 0.) Thus m - M = -M and m + M = M. So:

$\displaystyle v = \frac{m - M}{m + M}v_0 \approx \frac{- M}{M}v_0 = -v_0$

Since v is negative the golf ball has rebounded and is moving to the left after the collision with a speed of approximately 3 m/s.

-Dan