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A bullet of mass 10 g travelling horizon...

A bullet of mass 10 g travelling horizontally with a velocity of 150 m `s^(-1)` 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 he bullet.

Text Solution

Verified by Experts

`t = 0.03 s, m = 10g, v = 150 m//s`
`V = u + at`
`a = (-150)/(0.03) = -5000 m//s^(2)`
`V^(2) = u^(2) + 2as`
`S = (150 xx 150)/(10000)`
`S = 2*25` m
`F = ma`
` = (10)/(1000) xx - 5000 = - 500`
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Knowledge Check

  • A bullet of mass m moving with a horizontal velocity u strikes a stationary block of mass M suspended by a string of length L. If the bullet gets embedded, to what maximum angle, with vertical, the block would risc ?

    A
    `cos^(-1)[(m^(2)v^(2))/(2gL(M+m)^(2))]`
    B
    `tan^(-1)(m^(2)v^(2))/(2gL(M+m)^(2))`
    C
    `cos^(-1)[1-(m^(2)v^(2))/(2gL(M+m)^(2))]`
    D
    None of these

  • A bullet of mass a moving with velocity b strikes a large stationary block of wood of mass c, and remains embed in it, the final velocity of the system is :

    A
    `(b)/(c+b)`
    B
    `(a+b)/(c)a`
    C
    `(a)/(a+c)b`
    D
    `(a+b)/(a)b`

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