Physics, asked by wwwkartikeymishra104, 9 months ago

A bullet of mass 10g travelling horizontally with a velocity of 150m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the magnitude of the force exerted by the wooden block on the bullet.

Answers

Answered by anitajayara885161211
1

Answer:Given, Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg

Initial velocity of bullet, u = 150 m/s

Since bullet comes to rest, thus final velocity, v =0

Time, t = 0.03 s

Distance of penetration, i.e. Distance, covered (s)=?

Magnitude of force exerted by wooden block =?

We know that,

v = u + a t

0

=

150

m

s

1

+

a

×

0.03

s

150

m

/

s

=

a

×

0.03

s

a

=

150

m

/

s

0.03

  1. s

=

5000

m

s

2We know that,

s

=

u

t

+

1

2

a

t

2

s

=

150

m

/

s

×

0.03

s

+

1

2

(

5000

m

s

2

)

×

(

0.03

s

)

2

s

=

4.5

m

2500

m

s

2

×

0.0009

s

2

s

=

4.5

m

2.25

m

s

=

2.25

m

Magnitude of force exerted by wooden block

We know that, Force = mass x acceleration

Or,

F

=

0.01

k

g

×

5000

m

s

2

=

50

N

Therefore, Penetration of bullet in wooden block = 2.25 m

Force exerted by wooden block on bullet = – 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.

Answered by babukaleemsaabri
4

Answer:

50 N

Explaination

= given,mass=10g=0.010kg

v=150m/s

u=0m/s

t=0.03s

F=m×a

=m×(v-u)/t

=0.010kg×(150m/s-0m/s)/0.03s

=1kg×150m/s. /3s

= 50kgm/S2

= 50 Newton

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