A bullet of mass $10 g$ travelling horizontally with a velocity of $160 m s^{- 1}$ strikes a stationary wooden block and comes to rest in $0.02 s$ .The distance of penetration of the bullet into the block will be

1.20 m

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1.60 m

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2.00 m

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2.40 m

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Solution

The correct option is B $1.60 m$
u =160 m/s
t =0.02 s
v=0
v = u+at
0 = 160 +a(0.02)
$a = - 8000 m / s^{2}$
$v^{2} = u^{2} + 2 a s$
$s = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - {(160)}^{2}}{2 (- 8000)}$
$= \frac{- 25600}{- 16000} = 1.60 m$

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Similar questions

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 the bullet.

Q. Question 14
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 the bullet.

Q. A bullet of mass 10 g travlling 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 the bullet

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A bullet of mass 10 g travelling horizontally with a velocity of 160ms−1 strikes a stationary wooden block and comes to rest in 0.02 s.The distance of penetration of the bullet into the block will be

The correct option is B 1.60 mu =160 m/st =0.02 sv=0v = u+at0 = 160 +a(0.02)a=−8000m/s2v2=u2+2ass=v2−u22a=0−(160)22(−8000) =−25600−16000=1.60m

A bullet of mass $10 g$ travelling horizontally with a velocity of $160 m s^{- 1}$ strikes a stationary wooden block and comes to rest in $0.02 s$ .The distance of penetration of the bullet into the block will be

The correct option is B $1.60 m$
u =160 m/s
t =0.02 s
v=0
v = u+at
0 = 160 +a(0.02)
$a = - 8000 m / s^{2}$
$v^{2} = u^{2} + 2 a s$
$s = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - {(160)}^{2}}{2 (- 8000)}$
$= \frac{- 25600}{- 16000} = 1.60 m$