Question

As we all know Cousin Eddie of Christmas vacation loved to hunt and eat squirrels untill he learned that they were high in cholesterol. So now, he just plinks peices of 2x4 blocks off the back fence for fun, using his trusty .22 long rifle (.22LR). To prevent a through and through shot, he uses a 40 grain (2.60 grams; .0026kg) bullet having a muzzle velocity of 270 m/s.

He shoots into a 3/4 lb. (.340kg) peice of 2x4 sittin on the fence. The bullet imbeds itself 1.45 inches (36.8mm) into the wood. What was the final velocity of the bullet and wood block? Answer the following questions. (note Use total mass of both the bullet and block for final momentum and energy calculations.

1. P22i

2. P(2x4)

3. Ptoti

4. Ek22i

5. Ek(2x4)i

6.Ektoti

7.Vf

8.Ptotf

9.Ektotf

10. Does this show conservation of Momentum?

11. Does this show Conservation of Kinetic energy?

12. What was the average force of the bullet on the block?

Answer 1

The momentum before and after the collision will be conserved. The collision here is inelastic.

0.0026(270) + 0 = (0.0026 + 0.340)V

=> V = 1.936 m/s.

this is the velocity of the block and the bullet just after the bullet penetrates the wood.

Here, the kinetic energy is not conserved since the bullet and the block move together after the collision thereby implying that some energy is dissipated.

V = 1.936 m/s

final velocity of the bullet = V_f = 0 m/s

S = 36.8 mm = 36.8 x 10^-3 m

so, using V_f² = V² + 2aS gives:

a = - 50.926 m/s²

so, the average force of the bullet on the block is:

|F| = Ma = 0.34 x 50.926 = 17.315 N.