Question

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A spring whose spring constant k is is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block. The spring's maximum compression d is measured. Find an expression for the bullet's speed, Express your answer in terms of the variables m, M, k, d, and constant g.

Answer 1

Answer:

$v_o = \frac{1}{m}\sqrt{2(m + M)(\frac{1}{2}kd^2 + (m + M)gd)}$

Explanation:

First we will use momentum conservation for collision type of questions

So we will have

$m_1v_1 = (m_1 + m_2) v$

$m v_o = (m + M) v$

$v = \frac{mv_o}{m + M}$

now we can use energy conservation to find the maximum compression of spring

$W_s + W_g = 0 - \frac{1}{2}(m + M)v^2$

$- \frac{1}{2}kd^2 - (m + M)gd = - \frac{1}{2}(m + M)(\frac{mv_o}{m + M})^2$

so we have

$\frac{1}{2}kd^2 + (m + M)gd = \frac{1}{2(m + M)}m^2v_o^2$

$v_o = \frac{1}{m}\sqrt{2(m + M)(\frac{1}{2}kd^2 + (m + M)gd)}$

Answer 2

The expression for the bullet's speed when A block of mass M should be $v_0 = \frac{1}{m} \sqrt{ 2(m + M) (\frac{1 }{2}Kd^2 + (m + M)gd).$

Calculation of the expression:

Since

A block of mass M should be hung from the spring. A bullet of mass m should be fired vertically upward and into the bottom of the block. The spring's maximum compression d should be measured.

So based on this, the following formula should be used

$m_1v_1 = (m_1 + m_2)v\\\\mv_0 = (m + M)v\\\\v = mv_0\div m+ M$

Now

$Ws + Wg = 0 - 1\div 2(m + M)v^2\\\\-1\div 2kd^2 - (m + M)gd = -1\div 2(m + M) (mV_0\div m + M)2$

So, The expression for the bullet's speed should be $v_0 = \frac{1}{m} \sqrt{ 2(m + M) (\frac{1 }{2}Kd^2 + (m + M)gd).$

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