Respuesta :
To solve this problem we will use the theorems related to the conservation of momentum to determine the final speed of the system, once the bullet hits it. From there, we will calculate with the kinematic equations of linear motion, the height traveled with that speed. The conservation of momentum say us that,
[tex]m_1u_1+m_2u_2 = (m_1+m_2)v_f[/tex]
Where,
[tex]m_1 =[/tex] mass of bullet
[tex]m_2 =[/tex] mass of block of wood
[tex]u_{1,2} =[/tex] Initital velocity of bullet and block of wood respectively.
[tex]v_f =[/tex] Final velocity
Our values are given as,
[tex]m_1 = 1.40 Kg\\m_2 = 0.021 Kg\\u = 210 m/s[/tex]
Replacing we have that
[tex](1.40 + 0.021) * v_f = 0.021 * 210[/tex]
[tex]v_f = 3.103 m/s[/tex]
Note that the initial velocity of wood block is 0 because is at rest
Now using the kinematic equation of motion we have that
[tex]v_f^2-v_i^2 = 2gh[/tex]
As [tex]v_i = 0[/tex]
[tex]h = \frac{v_f^2}{2g}[/tex]
[tex]h = \frac{3.103^2}{2 * 9.8}[/tex]
[tex]h = 0.49 m[/tex]
Therefore the height reached by block after bullet embedding in it is 0.49 m
The block will rise 0.49 m.
By conservation of momentum, the momentum before the bullet gets embedded and after it must be same. After the bullet gets embedded the block and the bullet move together.
let the of block be M , mass of bullet be m, initial velocity of the bullet be u and the final velocity of the system be v. Then:
mu = (m + M) v
⇒ v = mu / (m + M)
v = (0.021×210) / (0021 + 210)
v = 3.1 m/s
To calculate the height to which the block rises:
at the maximum height, the final velocity of the block will be 0.
[tex]v^{2}=u^{2}-2gh[/tex]
here v = 0 and u = 3.1 m/s
[tex]h=\frac{u^{2} }{2g}[/tex]
[tex]= \frac{(3.1)^{2} }{2*9.8}[/tex]
h = 0.49 m
Learn more about conservation of momentum:
https://brainly.com/question/22257327
