## ashna 3 years ago A bullet of mass 20g strikes a block of mass 980g with a velocity V and is embedded with it . The block is in contact of a spring whose force-constant is 100N/M. After the collision the spring is compressed upto 10cm . find (a) the velocity of the block after collision (b) magnitude of the velocity v of the bullet (c) loss in KE due to the collision.

1. ashna

use v^2 - u^2 / 2as formula ?

2. ashna

then ?

3. imron07

Okay, while waiting for yahoo, let me give a clue:

4. ashna

okay

5. imron07

You cannot use v^2=blah-blah 2as formula because the acceleration here is not constant. Use conservation of energy instead: Kinetic energy+potential energy of the system is conserved.

6. ashna

okay

7. imron07

Did you get what i mean or just saying okay :)

8. imron07

For potential energy, use spring's potential. What is the kinetic energy + potential energy of the system after collition?

9. eashmore

imron07 is right, we need to track energy. Let's study the flow of energy. Let's ignore potential energy, we will assume everything occurs in a horizontal plane. Let's identify the components of the system. We have a bullet, block, and spring. First we fire the bullet. It has kinetic energy of$KE_b = {1 \over 2} m_b v^2$ Now, when the bullet impacts the block, we will assume all the energy is transferred. The kinetic energy of the bullet/block system is$KE_B = {1 \over 2} (m_b + m_B) V^2$(Big B is the block). Now, all that energy is transferred to the spring. The energy stored in a spring is$E_s = {1 \over 2} kx^2$where x is the displacement and k is the spring constant.

10. ashna

okay now i understand :)

11. ashna

what should i do for (a) ?

12. eashmore

You need to work backwards from what I presented. We know how much energy goes into the spring from what is given in the problem statement.

13. ashna

what is displacement here ?

14. ashna

@eashmore ?

15. eashmore

10 cm

16. ashna

first part E = 5000 J ?

17. eashmore

No. Remember to convert grams to kg. I made a mistake. The collision of the bullet and block is inelastic and therefore requires that we study the collision using conservation of momentum. $p_b = p'_b + p'_B$where ' indicates after the collision. They move at the same velocity so we have$m_b v_b = (m_b + m_B)V$ So, first find the energy stored in the spring. $E_s = {1 \over 2} k x^2$ This energy will be equal to the kinetic energy of the bullet/block system AFTER collision of the bullet. $E_s = KE_B \rightarrow KE_B = {1 \over 2} (m_b+m_B)V^2$We can find V from this. Then we can use the momentum equation to find the velocity of the bullet. Then find the energy from $E_l = KE_b - KE_B = {1 \over 2} m_bv^2 - {1 \over 2} (m_b + m_B)V^2$

18. ashna

okay .. i'll try ! Thanks for the help :)