How to Calculate Bullet Velocity After Penetrating a Block

[domingoleung]

Homework Statement

Block 1 (6 kg) is initially at rest on a rough and horizontal surface with coefficient of kinetic friction μk = 0.4. A bullet (0.05 kg) initially moving at speed u = 854 ms-1 firstly penetrates through block 1 which travels 0.1 m and comes to rest. Assume that there is no air resistance. Find the speed of the bullet vbu just after penetrating through block 1

Relevant Equations

Conservation of Momentum

Change in KE = Change in thermal energy
0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1
vblock = 0.885

By Conservation of Momentum,
(0.05)(854) = (0.05)*vbu + (6)(0.885)I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the block and the collision is not elastic

 

[PeroK]

Homework Statement:: Block 1 (6 kg) is initially at rest on a rough and horizontal surface with coefficient of kinetic friction μk = 0.4. A bullet (0.05 kg) initially moving at speed u = 854 ms-1 firstly penetrates through block 1 which travels 0.1 m and comes to rest. Assume that there is no air resistance. Find the speed of the bullet vbu just after penetrating through block 1
Relevant Equations:: Conservation of Momentum

Change in KE = Change in thermal energy
0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1
vblock = 0.885

By Conservation of Momentum,
(0.05)(854) = (0.05)*vbu + (6)(0.885)I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the block and the collision is not elastic

You need to include units in those numbers. What you have done so far looks about right. Why don't you finish things off?

 

[domingoleung]

You need to include units in those numbers. What you have done so far looks about right. Why don't you finish things off?

Yea thanks for reminding me. Vbu should be 747.8 m/s then :)

 

[PeroK]

Yea thanks for reminding me. Vbu should be 747.8 m/s then :)

Yes, although that's probably at least one significant figure too many.

 

[domingoleung]

Yes, although that's probably at least one significant figure too many.

Well, I don't quite understand why Δ KE equals to Δ Eth
Could you explain a bit?

 

[PeroK]

Well, I don't quite understand why Δ KE equals to Δ Eth
Could you explain a bit?

You know that momentum is conserved in the (nearly) instantaneous collision. The velocity of the block immediately after the collision tells you the velocity of the bullet immediately after the collision (which is what you want to calculate). And what you did calculate!

The velocity of the block after the collision can be calculated by how long it takes to stop under a known frictional force. Which, again, is what you did.

The last piece of the equation is the KE lost during the collision. You are not asked to calculate that, but you can if you want, using the velocities you have calculated.

The common theme in these problems is to use conservation of momentum but not conservation of kinetic energy. In general, you don't know how much kinetic energy was lost until you have completed the other calculations.

 

[domingoleung]

You know that momentum is conserved in the (nearly) instantaneous collision. The velocity of the block immediately after the collision tells you the velocity of the bullet immediately after the collision (which is what you want to calculate). And what you did calculate!

The velocity of the block after the collision can be calculated by how long it takes to stop under a known frictional force. Which, again, is what you did.

The last piece of the equation is the KE lost during the collision. You are not asked to calculate that, but you can if you want, using the velocities you have calculated.

The common theme in these problems is to use conservation of momentum but not conservation of kinetic energy. In general, you don't know how much kinetic energy was lost until you have completed the other calculations.

Thanks! <3 <3 <3