Question

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.

Answer 1

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

Answer 2

then ?

Answer 3

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

Answer 4

okay

Answer 5

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.

Answer 6

okay

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

okay now i understand :)

Answer 11

what should i do for (a) ?

Answer 12

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.

Answer 13

what is displacement here ?

Answer 14

@eashmore ?

Answer 15

10 cm

Answer 16

first part E = 5000 J ?

Answer 17

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 \[E_l = KE_b - KE_B = {1 \over 2} m_bv^2 - {1 \over 2} (m_b + m_B)V^2\]

Answer 18

okay .. i'll try !

Thanks for the help :)