In a ballistics demonstration, a police officer fires a bullet mass `50.0g` with speed `200ms^(-1)` on soft plywood of thickness 2.00cm. The bullet emerges only with `10%` of its initial kinetic energy. What is the emergent speed of the bullet ?

Options

A. `2sqrt10 m s^(-1)`
B. `20sqrt10 m s^(-1)`
C. `10sqrt2 m s^(-1)`
D. `10sqrt20 m s^(-1)`

Answer

B

Explanation

Here, m=50 g = `50 xx 10^(-3)` kg, `v_i=200 m s^(-1)`
initial kinetic energy of the bullet is
`K_i=1/2mv_i^2 =1/2xx(50xx10^(-3) kg)(200 m s^(-1))^2=1000 J`
Final kinetic energy of the bullet is
`K_f=10% K_i=10/100xx1000 J =100 J`
Let `v_f` be the emergent speed of the bullet. Then
`1/2 mv_f^2=K_f`
`v_f=sqrt((2K_f)/m)=sqrt((2xx100 J)/(50xx10^(-3) kg))=100sqrt(2/5) m s^(-1) =20sqrt10 ms^(-1)`