Question

You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged

Answer 1

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.