A bullet of mass 10 p travelling horizontally with a velocity of 150 m-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.
Answers
hello friend...!!
⇒ according to the question, we should calculate the distance penetrated by the bullet into the block
given that :
⇒ u = 150 m/s
⇒ v = 0
⇒ t = 0.03 sec
⇒ m = 10 g = 0,01 Kg
we know that , S = ( v² - u²) / 2a ------------( 1 )
we should calculate acceleration , v = u + at
a = ( v - u ) / t
a = ( 0 - 150 ) / 0.03
a = -5000 m/s² ---------------( 2 )
substituting equation ( 2 ) in ( 1 ) gives ,
S = { 0 - (150)² } / - ( 2 x 5000 )
S= 22500 / 10000
S = 2.25 m
therefore, the distance penetrated by the bullet is 2.25 m
to calculate force ,we know
F = m x a
⇒ F = 0.01 x 5000
⇒ F = 50 N
the force exerted by the wooden block on the bullet is 50 N
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Answer:
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