Question

A bullet of mass 10 g travelling horizontally with a velocity of 180 m/s strikes a stationary wooden block and comes to rest in 0.03 sec. Calculate the distance of penetration of the bullet in the block. Also calculate the magnitude of the force exerted by the block on the bullet.​

Answer 1

 Initial velocity of the bullet, u=150  ms

−1

Final velocity of the bullet, v=0

Mass of the bullet, m=10 g=0.01 kg

Time taken by the bullet to come to rest, t=0.03 s

Let distance of penetration be s

v=u+at

0=150+a(0.03)  

⟹a=−5000 ms

−2

    (- sign shows retardation)

v

2

=u

2

+2as

0=150

2

+2×(−5000)×s

s=2.25 m

Magnitude of the force exerted by the wooden block, ∣F∣=m∣a∣

∣F∣=0.01×5000=50 N                                                                                                                                                        

bro pls give five star

Answer 2

Solution : -

$\sf{Here,} \\ \\ \sf{m = 10} \\ \\ \sf{g = 0.01 kg}, \\ \\ \sf{u = 150 m s^{ - 1}} , \\ \\ \sf{v = 0}, \\ \\ \sf{t = 0.03 s} \\ \\ \implies \sf \boxed{\green{a=\dfrac{v - u}{t}}} \\ \\ \implies \sf{ \frac{0 - 150}{0.03}} \\ \\ \implies \sf{ - 5000 \: ms^{ - 2}} \\ \\ \sf{The \: distance \: of \:penetration \: of \: the \: bullet into \: the \: block} \\ \\ \implies\sf\boxed{\red{s = ut + \frac{1}{2} at^{2}}} \\ \\ \implies\sf{ 150 \: \times \: 0.03+ \frac{1}{2}x (-5,000) \times (0.03)^{2}} \\ \\ \implies \sf{ 4.5 - 2.25} \\ \\ \implies\sf{2.25 m \: Ans.} \\ \\ \sf{The \: magnitude \: of \: the \: force \: exerted \: by \: the \: wooden \: block \: on \: the \: bullet} \\ \\ \implies \sf{ ma = 0.01 \times 5,000} \\ \\ \implies \sf{50 N. \: Ans}$