Answer:
4.88505 m/s
1.21629 m
Explanation:
[tex]m_1[/tex] = Mass of bullet = 2.5 g
[tex]m_2[/tex] = Mass of block = 215 g
[tex]v_1[/tex] = Velocity of bullet = 425 m/s
[tex]v_2[/tex] = Velocity of block
Here, the linear momentum is conserved
[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{2.5\times 10^{-3}\times 425+0.215\times 0}{2.5\times 10^{-3}+0.215}\\\Rightarrow v=4.88505\ m/s[/tex]
The speed of the combination of mass is 4.88505 m/s
The energy in the system is conserved
[tex]\dfrac{1}{2}(m_1+m_2)v^2=(m_1+m_2)gh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{4.88505^2}{2\times 9.81}\\\Rightarrow h=1.21629\ m[/tex]
The maximum height the combined mass will reach is 1.21629 m
Answer:
Explanation:
mass of bullet, m = 2.5 g
initial velocity of bullet, u = 425 m/s
mass of block, M = 215 g
(a) Let the speed of bullet block system is v
Use the conservation of momentum
m x u + M x 0 = (M + m) x v
2.5 x 425 = ( 215 + 2.5) x v
v = 4.89 m/s
(b) Let it rise upto height h
Use conservation of energy
Kinetic energy after collision = Potential energy
1/2 (M + m) v² = ( M+ m) x gx h
0.5 x 4.89 x 4.89 = 9.8 x h
h = 1.22 m
