Answer:
[tex]K=250.11 J[/tex]
Explanation:
Mass of the bullet (m) = 0.013 kg
Mass of the block (m_1) = 9 kg
Height lift of the block (x) = 7.0 mm
Consider the velocity of the bullet and the block after the collision is "V".
Therefore the energy after the collision
[tex]E_1=0.5(m+m_1)*v^2[/tex]
Now this energy will be converted into the potential energy when the block lifts, therefore applying energy conservation
[tex]E_1=(m+m_1)*gh[/tex]
[tex]0.5*(m+m_1)*v^2=(m+m_1)g*h[/tex]
[tex]v^2=2*9.8*7x10^{-3}[/tex]
[tex]v_1=0.37m/s[/tex]
Now consider the initial velocity of the bullet is "v".
Therefore applying conservation of momentum
[tex]m*v=(m+m_1)*v_1[/tex]
[tex]0.017kg*v=(0.013kg+9kg)*0.37m/s[/tex]
[tex]v=196.16m/s[/tex]
Now the kinetic energy of the bullet
[tex]K=\frac{1}{2}*m*V^2[/tex]
[tex]K=\frac{1}{2}*0.013kg*(196.16)^2[/tex]
[tex]K=250.11 J[/tex]
The initial kinetic energy of the bullet before the collision is 6.12 x 10⁵ J.
The given parameters;
The acceleration of the bullet-block system is calculated as follows;
s = ut + ¹/₂at²
0.007 = 0 + 0.5(0.001)²a
0.007 = 5 x 10⁻⁷ a
[tex]a = \frac{0.007}{5\times 10^{-7}} \\\\a = 14000 \ m/s^2[/tex]
The final velocity of the bullet block system is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2(14000)(0.007)\\\\v^2 = 196 \\\\v= \sqrt{196} \\\\v = 14 \ m/s[/tex]
The initial velocity of the bullet is calculated as follows;
[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.013u_1 + 9(0) = 14(9 + 0.013)\\\\0.013 u_1 = 126.18\\\\u_1 = 9,706.2 \ m/s^2[/tex]
The initial kinetic energy of the bullet is calculated as follows;
[tex]K.E_i = \frac{1}{2} mu_i^2\\\\K.E_i = \frac{1}{2} \times 0.013 \times (9,706.2)^2\\\\K.E_i = 6.12\times 10^5 \ J[/tex]
Thus, the initial kinetic energy of the bullet before the collision is 6.12 x 10⁵ J.
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