Answer:
It was not fired from the client's gun because the chair slid only 3 centimeters . If it had been fired from the client's gun the chair would slid 25.82 centimeters.
Explanation:
According to the law of conservation of momentum the momentum of the system before collision must be equal to the momentum of the system after the collision.
M1u1= m2u2
Let M1 = mass of the chair = 20kg
m2= mass of the bullet= 10g= 0.001kg
u1= velocity of the chair before collision = zero m/s
u2 = velocity of the bullet before collision = zero m/s
v1= velocity of the chair after collision = ? m/s
v2 = velocity of the bullet after collision = 450 m/s
After collision their velocities change from u1 to v1 and u2 to v2 so
M1v1= m2v2
v1= m2v2/M1
v1= 0.01 *450/ 20= 0.225 m/s
Now according to the law of conservation of energy the energy of the system before collision must be equal to the energy of the system after the collision.
The energy of the chair after the bullet is hit is
KE of the chair + KE of the bullet= 1/2 (M)(v1)²+ 1/2 m(v2)²=
1/2 ( 20) (0.225 )² + 1/2 (0.01) (450)²
= 0.50625 + 1012.5= 1013.00625 Joules
Frictional force = Coefficient of kinetic force of wood on wood ( M+m) g
= 0.2* ( 20.01) 9.8= 39.2196 N
Work done by friction = frictional force * distance
If law of conservation of energy is applied the KE must be equal to the work done
KE = W
W= f*d
KE= F*d
d = KE/f= 1013.00625/ 39.2196= 25.82 cm
The chair did not move 25.82 cm .
It only moved 3 centimeter.
Hence the bullet fired was not from the client's gun.
