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