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A 12.0 g bullet was fired horizontally into a 1 kg block of wood. The bullet initially had a speed of 250 m/s. The block of wood was hanging from a 2 m long piece of (massless) string. After the collision the block/bullet combined object swings upward on the string. Find the height the block/bullet combined object rises.

a. 0.66m
b. 0.45m
c. 0.12 m
d. 0.35 m
e. 0.27m

Respuesta :

Answer:

So height though which combination of block bullet rises is 0.45 m

Explanation:

We have given  mass of the bullet  [tex]m_1=12gram=0.012kg[/tex]

Velocity of the bullet [tex]v_1=250m/sec[/tex]

Mass of block of wood [tex]m_2=1kg[/tex]

Block is at rest so [tex]v_2=0m/sec[/tex]

From conservation of momentum.

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]0.012\times 250+0\times 0=(1+0.012)v[/tex]

[tex]v=2.964m/sec[/tex]

From third equation of motion

[tex]h=\frac{v^2}{2g}=\frac{2.964^2}{2\times .8}=0.45m[/tex]

So option (b) will be the correct answer.

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