Question

A bullet of mass 10g travelling horizontally with a velocity of 150m^-1 strikes a stationary wooden block and comes to rest in 0.03s.calculate the magnitude of the of the force exerted by the wooden block on the bullet?Also calculate the distance of penetration of the bullet on the block?

Answer 1

Answer:

Given:

Initial velocity (u)= 150m/s

Final velocity (v) = 0m/s

Time taken (t) = 0.03s

Mass(m) = 10g

$= \frac{10}{1000}$

$= 0.01kg$

To find :

a) Force exerted by the wooden block on the bullet.

b) The distance of penetration of the bullet on the block.

Formulae:

$a)f = m \times a$

$b)v {}^{2} = u {}^{2} + 2as$

Solution :

a) According to the first equation of motion,

$v = u + at$

Acceleration of the bullet,

$0 = 150 + a(0.03)$

$a = \frac{ - 150}{0.03}$

$a = - 5000m s {}^{ - 2}$

Negative sign indicates that the velocity of the bullet is decreasing.

According to the third equation of motion,

$v {}^{2} = u {}^{2} + 2as$

$0 = (150)2 + 2( - 5000)$

$= \frac{22500}{10000}$

$= 2.25m$

Hence, the distance of penetration of the bullet into the block is 2.25 m.

b)

$force = mass \times acceleration$

$f = 0.01 \times ( - 5000)$

$f = 50newton$

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