The bullet is 20 grams. The wood block is 400 grams. It is 10 cm high and 20 cm

Question

The bullet is 20 grams. The wood block is 400 grams. It is 10 cm high and 20 cm wide. The thickness will not be necessary. The center of mass of the block travels a distance of 1 meter after the collision. The bullet enters the block 5 cm from the center in case 2. Let’s assume that it takes 10 J of energy to penetrate the block of wood by 1 mm

b. Calculate the energy loss difference

c. Calculate the penetration depth difference.

CASE 1 CASE 2 wooden block width 20 cm rotating block and bullet zero velocity wooden block height 10 cm block elevation 1 m block elevation 1 m wooden block 400 g bullet 20 g entry point 5 cm off center

Explanation / Answer

In Case 1:

Conserving momentum just after and before the collision :

m*u + 0 = (M+m)*v

So, 20*u = (400+20)*v

So, v = u/21

For the height reached ,

h = v^2/2g = (u/21)^2/(2*g)

So, 1 = (u/21)^2/(2*9.8)

So, u = 92.97 m/s

So, energy loss, U = 0.5*mu^2 - 0.5*(m+M)*v^2

= 0.5*0.02*92.97^2 - 0.5*(0.420)*(92.97/21)^2

= 82.3 J

So, penetration depth, d = 82.3/10 = 8.23 mm

For case 2:

Conserving angular momentum with respect to the center of mass of the block:

m*u*x = (M/12)*(0.1^2 + 0.2^2)*W <------ W = angular speed

So, 20*u*0.05 = (400/12)*(0.1^2 +0.2^2)*W ------ (1)

Conserving linear momentum :

20*u = (420)*v ----------- (2)

Conserving energy after collision:

0.5*20*v^2 = 0.5*((400/12)*(0.1^2 +0.2^2) + 20*0.05^2)*W^2 + (420)*9.8*1 ---------(1)

Solving the three equations,

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