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14. A bullet of mass 10 g travelling horizontally with a velocity of
150 m s-l strikes a stationary wooden block and comes to rest
in 0.03 s. Calculate the distance of penetration of the bullet
into the block. Also calculate the magnitude of the force exerted
by the wooden block on the bullet.​

Respuesta :

Answer:

2.25 m

50 N

Explanation:

Given :

[tex]m=10 g=0.01 kg[/tex]

Initial velocity =U= [tex]150 \ ms^{-1}[/tex]

[tex]Time=t=0.03\ s[/tex]

Final velocity =v=0

We know that

[tex]V=U + at[/tex]

[tex]a\ =\ \frac{V-U}{t}[/tex]

Putting the value of V,U and t in the previous equation we get

[tex]a=\frac{0-150}{0.03} \ \\a=-5000 ms^{-2}[/tex]

According the second law of motion

[tex]s=Ut+\frac{1}{2} *\ at^{2}[/tex]

Putting the value of U,t and a we get

[tex]s=150\ *\ 0.03\ +\frac{1}{2} \ *(-5000)\ * 0.03\ *0.03\\s=2.25\ m[/tex]

We know that

[tex]F=mass * acceleration\ \\F=0.01 * 5000\ \\F=50 N[/tex]

  • Distance of penetration is 2.25 m
  • The magnitude of the force Is :50 N

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