A 4.0 g bullet traveling horizontally with a speed of 400 m/s, is fired into a w

Question

A 4.0 g bullet traveling horizontally with a speed of 400 m/s, is fired into a wooden block with mass 0.8 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190m/s. What is the speed of the wooden block after the collision? Calculate the total kinetic energy before and after collision and check if it is conserved or not. Is the collision elaslic, inelastic or totally inelastic ?. Explain. Find the average net force on the bullet by assuming an interaction time of 0.010 s.

Explanation / Answer

mass of the bullet, m1=4g

speed of the bullet, u1=400 m/sec

mass of the block, m2=0.8 kg and u2=0

after collision,

speed of the bullet, v1=190 m/sec

speed of the block is v2

now,

1)

use law of conservation mometum,

m1*u1+m2*u2=m1*v1+m2*v2

(4*10^-3)*400+(0.8*0)= (4*10^-3)*190+(0.8*v2)

====> v2=1.05 m/sec

speed of the block after collision, v2=1.05 m/sec

2)

before collision,

total K.Ei=1/2*m1*u1^2 + 1/2*m2*u2^2

=1/2*(4*10^-3)*400^2 + 0

=320 m/sec

now,

after collision,

total K.Ef=1/2*m1*v1^2 + 1/2*m2*v2^2

=1/2*(4*10^-3)*190^2 + 1/2*0.8*1.05^2

=72.641 m/sec

here,

K.Ei is not equals to K.Ef

energy is not conserved

3)

here,

in this collision,

momenetum is conserved

but,

K.E is not conserved

hence, it is a inelastic collision

4)

impulse, I=Pi-Pf

I=m1*u1-m*u2

I=m1*(u1-u2)

I=(4*10^-3)*(400-190)

I=0.84 kg.m/sec

but,

I=Favg*t

0.84=Favg*0.01

====> Favg=84 N

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