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Using second law of motion, derive the r...

Using second law of motion, derive the relation between force and acceleration. A bullet of `10g` strikes a sand-bag at a speed of `10^(3)ms^(-1)` and gets embedded after travelling `5cm`.Calculate
(i) the resistive force exerted by the sand on the bullet (ii) the time taken by the bullet to come to rest.

Text Solution

Verified by Experts

The required relation is `F=ma`
Mass of bullet, `m=10g=10^(-2)kg`
speed of bullet, `v=10^(3)ms^(-1)`
distance travelled, `s=5cm=5xx10^(-2)m`
As work done= `K.E` bullet
`:. Fxxs=1/2mv^(2)`
`F=(mv^(2))/(2s)=(10^(-2)(10^(3))^(2))/(2xx5xx10^(-2))=10^(5)N`
From `v=u+at`
`0=10^(3)+((-10^(5))/(10^(-2)))xxt, t=10^(3)/(10^(7))=10^(-4)s`
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Knowledge Check

  • A bullet of mass 40g moving with a speed of 90ms^(-1) enters a heavy wooden block and is stopped after a direction of 60cm. The average resistive force exered by the block on the bullet is

    A
    180N
    B
    220N
    C
    270N
    D
    320N

  • A 10.0-g bullet traveling horizontally at 755 m/s strikes a stationary target and stops after penetrating 14.5 cm into the target. What is the average force of the target on the bullet ?

    A
    `1.97xx10^(4)`N
    B
    `6.26xx10^(3)`N
    C
    `2.07xx10^(5)`N
    D
    `3.13xx10^(4)N`