Respuesta :
Answer:
The height raised by the blcok is 2.5 mm.
Explanation:
Given that,
Mass of the bullet, [tex]m_1=3\ g=0.003\ kg[/tex]
Initial speed of the bullet, [tex]u_1=750\ m/s[/tex]
Mass of the block of wood, [tex]m_2=10\ kg[/tex]
It was at rest, initial speed of the blockof wood, [tex]u_2=0[/tex]
The block swings in an arc, rising a height h from its lowest position. The momentum remains consered. Let V is the velocity of the block +bullet system together. So,
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{0.003\times 750+0}{(0.003+10)}\\\\V=0.224\ m/s[/tex]
Now using the conservation of energy to find the height h raised by the blck. So,
[tex]\dfrac{1}{2}mV^2=mgh\\\\h=\dfrac{V^2}{2g}\\\\h=\dfrac{(0.224)^2}{2\times 10}\\\\h=0.0025\ m\\\\h=2.5\ mm[/tex]
So, the height raised by the blcok is 2.5 mm.
Answer:
2.58 mm
Explanation:
mass of bullet, m = 3 g = 0.003 kg
mass of block M = 10 kg
initial velocity of bullet, u = 750 m/s
Let they move together after collision with velocity v.
Use conservation of momentum
m x u + 0 + ( m + M) x v
0.003 x 750 = ( 10 + 0.003) x v
v = 0.225 m/s
Let the both reach at height h.
use conservation of energy
Kinetic energy at the bottom = Potential energy at height
0.5 x ( M + m) x v² = ( M + m) x g x h
0.5 x 0.225 x 0.225 = 9.8 x h
h = 2.58 x 10^-3 m
h = 2.58 mm
