Question

Elastic Collisions: "A charging bull elephant with a mass of 5400kg comes directly toward you with a speed of 4.30 m/s. You toss a 0.150 kg ball at the elephant with a speed of 8.11 m/s. "
(a) When the ball bounces back toward you, what is its speed?
(b) How do you account for the fact that the ball's kinetic energy has increased?

**The answer in the back of the book for (a) is 17 m/s.**

Answer 1

I got 16.7 m/s perhaps only approximate answer.

Started with conservation of momentum M v1 - m v2 = M v1' + m v2'

where primes mean after collision
Then used conservation of energy

1/2 M v1^2 + 1/2 m v2^2 = 1/2 M v1'^2 + 1/2 m v2'^2

The put the Ms on left side of each equation and the ms on the right.

Divided the momentum equation into the energy equation,
noting (a^2-b^2)=(a+b)(a-b)

Assumed M so much larger than m so that v1+v1' = 2 v1,

got 2(4.30) = v2' - 8.11

v2' = 16.7.

The extra energy comes from the motion of the beast, slightly slowed.

Answer 2

NOTE! This way definitely works, but it's really, really ugly :P

You can do it without assuming infinite mass of the elephant (although granted it doesn't change the answer much)
Starting with your conservation of momentum
\[p_i=p_f\]
\[Mv_1 - mv_2 = Mv_1' + mv_2'\]
\[\qquad \qquad \qquad A= (Mv_1-mv_2) \quad \leftarrow \text{do this to make our lives easier!}\]
\[ A = Mv_1' - mv_2'\]
\[Mv_1' = A-mv_2'\]
\[(Mv_1')^2 = (A-mv_2')^2\]
\[Mv_1^{' \ 2}=\frac{(A-mv_2')^2}{M}\]

Then in your conservation of energy (just drop all of the halves off of the bat since every term has them)
\[Mv_1^2 + mv_2^2 = Mv_1^{' \ 2} + mv_2^{' \ 2}\]
sub in our M(v1')^2 from above
\[Mv_1^2 + mv_2^2 = \frac{(A-mv_2')^2}{M} + mv_2^{' \ 2}\]
So now solve or v2' - the velocity of the ball
\[ Mm(v_2')^2 +A^2 - 2Amv_2' + (mv_2')^2= MMv_1^2 + Mmv_2^2\]
ick!!
\[Mmv_2^{' \ 2}+(mv_2')^2-2Amv_2'+A^2-(MMv_1^2 + Mmv_2^2)=0\]
\[v_2^{' \ 2}(Mm+m^2)-2Amv_2'+A^2-(MMv_1^2 + Mmv_2^2)=0\]
triple ick. We can see that it's a quadratic on v_2' however, so
\[ (v_2')^2 \Big( Mm+m^2 \Big) + v_2' \Big( -2Am \Big) + \Big( A^2 - (MMv_1^2 + Mmv_2^2) \Big) = 0\]

to use in

\[v_2' = \frac{-b±\sqrt{b^2-4ac}}{2a}\]

quadruple ick

\[a = Mm+m^2 = ((5400kg)(.150kg)+(.150kg)^2) \approx 795kg^2 \]
\[b = (-2Am) = 2(Mv_1' + mv_2')(m)
\\ = 2((5400kg)(4.3m/s)-(.15kg)(8.11m/.s))(.15kg)
\\ b \approx -6966 kg^2 m/s\]
\[c = A^2-(MMv_1^2 + Mmv_2^2)
\\ =((5400kg)(4.3m/s)-(.15kg)(8.11m/.s))^2-(5400kg)^2(4.30m/s)^2
\\ \qquad - (5400kg)(.15kg)(8.11m/s)^2
\\ c \approx -109768 kg^2 m^2/s^2\]

Just yuck!

then we get

\[v_2' = \frac{-b±\sqrt{b^2-4ac}}{2a}
\\ \
\\ =\frac{6966 kg^2m/s ± \sqrt{(-6966 kg^2m/s)^2-4(795kg^2)(-109768 kg^2m^2/s^2))}}{2(795 kg^2)}\]
giving two solutions

\[v_2'=-8.16 m/s\]
\[v_2'= 16.92 m/s\]

We want the positive one, because the elephant was originally traveling in the positive direction, so the ball will be, too.


Methinks the other way was better...........

Answer 3

I'm sorry, this is way less than helpful :P