Question

A 7.00 g bullet, when fired from a gun into a 0.80 kg block of wood held in a vise, penetrates the block to a depth of 7.00 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7.00 g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case?
(............) cm

Answer 1

When First Bullet Hit

The Kinetic Energy of the Bullet = Work Done by Frictional Force in the Block

0.5*m*v^2 = FS

F = (0.5*7*10^-3*v^2)/(0.07)

Therefore

F = 0.05v^2 ---------------------------(1)

When Second Bullet Hit

Momentum Remains Conserved

Therefore

mv = (M + m + m)*V

(7*10^-3)v = (0.80 + 2*7*10^-3)*V

V = 8.6*10^-3 v --------------------------(2)

Also

The Kinetic Energy of the Bullet = Work Done by Frictional Force in the Block

F*S' = 0.5*(M + m + m)*V^2

From Eq(1) , put the Value of F and From Eq(2) put the value of V we get

0.05v^2*S' = 0.5*(0.8 + 2*7*10^-3)*((8.6*10^-3)v)^2

S' = 0.00060 m

= 0.060 cm

Answer 2

Answer 3

depth of penetration D is proportional to energy loss:

energy loss = force * distance

In the former case energy loss in collision is equal to initial kinetic energy of the bullet:

force * distance1 = ΔE1

force * distance1 = 1/2 mv²

In the latter case of frictionless surface, linear momentum of bullet + block is conserved

Energy of center of mass is

Ecm = 1/2 p²/(M+m)

Ecm = 1/2 (mv)²/(M+m)

Ecm = 1/2 mv² * m/(M + m)

Again, depth of penetration in this case is proportional to the loss of energy:

force * distance2 = ΔE2

force * distance2 = K - Ecm

force * distance2 = 1/2 mv² - 1/2 mv² * m/(M + m)

force * distance2 = 1/2 mv² * M/(M + m)

so,

D2/D1 = M/(M + m)

D2 = D1 * M/(M + m)

D2 = 7.4 * 1000/1007

D2 = 7.34856 cm

depth the bullet will penetrate = 7.34856 cm