Answer:
0.358326032852 kg
Explanation:
[tex]m_1=m_b[/tex] = Mass of bullet = 11.5 g
[tex]v_1=v_b[/tex] = Velocity of bullet = 265 m/s
x = Displacement of spring = 35 cm
v = Velocity of the system
[tex]m_2[/tex] = Mass of the block
As the momentum of the system is conserved we have
[tex]m_1u_1=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1}{m_1+m_2}[/tex]
As the energy in the system is conserved we have
[tex]\dfrac{1}{2}(m_1+m_2)v^2=\dfrac{1}{2}kx^2\\\Rightarrow (m_1+m_2)\left(\dfrac{m_1u_1}{m_1+m_2}\right)^2=kx^2\\\Rightarrow m_2=\dfrac{m_1^2u_1^2}{kx^2}-m_1\\\Rightarrow m_2=\dfrac{0.0115^2\times 265^2}{205\times 0.35^2}-0.0115\\\Rightarrow m_2=0.358326032852\ kg[/tex]
The mass of the block is 0.358326032852 kg
Answer:
Explanation:
mass of bullet, mb = 11.5 g
velocity of bullet, vb = 265 m/s
Spring constant, K = 205 N/m
compression in spring, Δd = 35 cm = 0.35 m
Let m be the mass of wooden block.
Let v is the velocity of wooden block.
Use conservation of momentum
mb x vb = ( m + mb) x v
0.0115 x 265 = ( m + 0.0115) x v
[tex]v = \frac{3.05}{m + 0.0115}[/tex] ... (1)
Use conservation of energy
[tex]\frac{1}{2}K \times \Delta d^{2}=\frac{1}{2}\left ( m + m_{b} \right )v^{2}[/tex]
[tex]205\times 0.35\times 0.35=\left ( m + m_{b} \right )\times \left ( \frac{3.05}{m + m_{b}} \right )^{2}[/tex]
[tex]25.1125\times (m + 0.0115)=9.3025[/tex]
m = 0.36 kg
