MHB Bullet imbedding in a block on a frictionless surface

Click For Summary
In an inelastic collision, a 10g bullet traveling at 1000 m/s strikes a 100g block at rest, resulting in a combined velocity calculated using conservation of momentum. The formula yields a final velocity of approximately 90.91 m/s for the bullet-block system. Conservation of energy cannot be applied here, as kinetic energy is not conserved in inelastic collisions. The lack of information regarding the forces during the collision further restricts the use of Newton's Second Law. Therefore, conservation of momentum is the only applicable principle for this scenario.
Dustinsfl
Messages
2,217
Reaction score
5
A 10g bullet with v = 1000 m/s strikes a 100g block which is at rest. What is their combined velocity? Can this be done with conservation of energy? momemtum?

Here we have inelastic collision so by conservation of momentum
\[
10\cdot 1000 + 0 = 110\cdot v_f\Rightarrow v_f = \frac{1000}{11}m/s.
\]
Can this be done with CoE? If so, how?
 
Mathematics news on Phys.org
Re: bullet imbedding in a block on a frictionless surface

dwsmith said:
A 10g bullet with v = 1000 m/s strikes a 100g block which is at rest. What is their combined velocity? Can this be done with conservation of energy? momemtum?

Here we have inelastic collision so by conservation of momentum
\[
10\cdot 1000 + 0 = 110\cdot v_f\Rightarrow v_f = \frac{1000}{11}m/s.
\]
Can this be done with CoE? If so, how?

No, it cannot, because the collision is inelastic. An elastic collision, by definition, is a collision in which kinetic energy is conserved. With an inelastic collision, it is not conserved. Moreover, you're given no information about the force during the collision, so you can't use Newton's Second Law, or a more general conservation of energy approach. Conservation of Momentum is all that's available to you.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
55
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K