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A 0.00545-kg bullet is fired straight up at a falling wooden block that has a mass of 3.08 kg. The bullet has a speed of 720 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

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Answer:

0.06s

Explanation:

We can solve a problem with collision using the principle of conservation of momentum, where

[tex]m_1u_1+m_2u_2 = (m_1+m_2)v[/tex]

Where v is the velocity after the collision,

Our values are given by,

[tex]m_1=0.00545\\u_1=  720m/s\\m_2= 3.08kg\\u_2=v[/tex]

Replacing we have,

[tex](0.00545)(720)+(3.08)(-v)=(3.08+0.00545)v[/tex]

[tex]3.924-3.08V = 3.08V +0.00545V[/tex]

[tex]V = \frac{3.924}{6.165}[/tex]

[tex]V = 0.6364m/s[/tex]

Note= Velocity block is given negative becouse V=-gt before collision, i.e, the direction changes.

For time we can use the equation of gravity, solving for t

[tex]g=\frac{v}{t}[/tex]

[tex]t= \frac{v}{g}[/tex]

[tex]t= 0.6364/0.8[/tex]

[tex]t= 0.06s[/tex]

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