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A bullet Is fired into a block of wood sitting on a block of ice. The bullet has an initial velocity of 800 m/s and a mass of .007 kg. The wooden block has a mass of 1.3 kg is and is initially at rest. The bullet remains in bedded in the blackboard afterward.

Assuming that momentum is conserved what is the velocity of the block of wood and bullet after the collision? Round to the nearest hundredths place.

What is the magnitude of the impulse that axle a block of wood in this process? Round to the nearest hundredths place.

Respuesta :

curiousmuktar

Answer:

43.1

Explanation:

Mv=Mv

0.07*800=1.3*V

V=56/1.3

V=43.07

then 43.1

Answer:

[tex]v = 4.28 m/s[/tex]

[tex]Impulse = 5.57 kg m/s[/tex]

Explanation:

Here we can say that bullet + block system is an isolated system and there is no external force on it

So Net momentum of bullet + block system will remains conserved

So we will say

[tex]m v_o = (M + m) v[/tex]

so we will have

[tex]v = \frac{m}{M + m} v_o[/tex]

[tex]v = \frac{0.007 (800)}{1.3 + 0.007}[/tex]

[tex]v = 4.28 m/s[/tex]

Also in order to find the impulse on the block we know that

impulse = change in the momentum of the block

[tex]Impulse = m(v_f - v_i)[/tex]

[tex]Impulse = 1.3(4.28 - 0)[/tex]

[tex]Impulse = 5.57 kg m/s[/tex]

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