Hammer striking an anvil with a velocity of 50ft/sec

[Dustinsfl]

Homework Statement

A hammer strikes an anvil with a velocity of 50ft/sec. The hammer weights 12 lb and the anvil weighs 100 lb. The anvil is supported on four springs with k = 100 lb/in. Find the motion if
(a) the hammer stays in contact with the anvil
(b) the hammer doesn't remain in contatct

Homework Equations

$1 lb = 4.45 N$
$1 m = 3.28ft$
$k_{eq} = 4k = 400\cdot 4.45\cdot 12\cdot 3.28 = 70060.8$ N/m
$W = mg$ so $m_h = 12*4.45/9.8 = 5.45$ kg and $m_a = 100*4.45/9.8 = 45.41$ kg
$\dot{x}(0) = 50/3.28 = 15.24$ m/s

The Attempt at a Solution

(a)
$$
M\ddot{x} + k_{eq}x = 0
$$
where $M = m_h + m_a = 50.86$ kg.
$$
\ddot{x} + \omega_n^2x = 0
$$
where $\omega_n^2 = \frac{70060.8}{50.86} = 1377.52$
Let $x(t) = A\cos(\omega_nt) + B\sin(\omega_nt)$. Then $\ddot{x} = -A\omega_n^2\cos(\omega_nt) - B\omega_n^2\sin(\omega_nt)$.
$$
\cos(\omega_nt)[-A\omega_n^2 + A\omega_n^2] + \sin(\omega_nt)[-B\omega_n^2 + B\omega_n^2] = 0
$$
The coefficient zero out so this can't be correct. The RHS has to be zero since there is no driving force.

 

[haruspex]

Leave out all the numbers and conversions for now, and just do everything symbolically. Hammer mass m moving vertically down at speed u strikes anvil of mass M supported on four springs of coefficient k. What is the speed of the anvil immediately after impact?

 

[Dustinsfl]

Leave out all the numbers and conversions for now, and just do everything symbolically. Hammer mass m moving vertically down at speed u strikes anvil of mass M supported on four springs of coefficient k. What is the speed of the anvil immediately after impact?

I have been thinking about this. Should the initial velocity for the inelastic collision be found as
$$
m_h(50)^2 = v^2(m_h + m_a)
$$
Then v would be $\dot{x}(0) = v$?

For part b, since the hammer comes off immediately, that would be view as a delta spike and $\dot{x}(0) = 50$ the velocity of the hammer, correct?

 

[haruspex]

I have been thinking about this. Should the initial velocity for the inelastic collision be found as
$$
m_h(50)^2 = v^2(m_h + m_a)
$$
Then v would be $\dot{x}(0) = v$?

Since it is an inelastic collision, work is not conserved. What is conserved?

 

[rude man]

Try instead using momentum conservation for both cases, then relate decreasing kinetic energy to increasing potential energy after contact.