- #1

hairey94

- 15

- 2

- Homework Statement
- A 12 g bullet shot vertically into a 5 kg block and lifting it upwards to a maximum height of 4 mm. The bullet traveled for 1 ms in the block before stopping completely. Calculate the depth of the bullet embedded in the block.

- Relevant Equations
- The answer should be 0.06 m but I can't get it.

Linear Motion Equation to get the common velocity of the block and bullet just after collision:

v

I got u = 0.28 m/s

Then I calculate the impulse of the block using formula J = mv - mu, where v=0 and u=0.28 m/s and I got J = 1.4 N s

The last attempt is to use the conservation of linear momentum since the collision is inelastic and kinetic energy is not conserved:

mv' = J where v' is the velocity of the bullet just before the collision

I got v' = 116.95 m/s and this value seems reasonable and I continue using linear motion equation:

s = v't + (1/2)gt

but at the end the answer for s is not 0.06 m. Please let me know where I did wrong.

v

^{2}=u^{2}+2gs, I set v=0 at max height of s = 0.004 m and g = -9.81 m s^{-2}I got u = 0.28 m/s

Then I calculate the impulse of the block using formula J = mv - mu, where v=0 and u=0.28 m/s and I got J = 1.4 N s

The last attempt is to use the conservation of linear momentum since the collision is inelastic and kinetic energy is not conserved:

mv' = J where v' is the velocity of the bullet just before the collision

I got v' = 116.95 m/s and this value seems reasonable and I continue using linear motion equation:

s = v't + (1/2)gt

^{2}where the t = 0.001 s and g = -9.81 m s^{-2}but at the end the answer for s is not 0.06 m. Please let me know where I did wrong.