3. Linear Momentum/Work and Energy: (7 points) A bullet of mass 15 grams is fire

Question

3. Linear Momentum/Work and Energy: (7 points) A bullet of mass 15 grams is fired into a block of mass 985 grams, which is attached to an uncompressed spring of force constant A 1000 The velocity of the bullet is parallel to the long dimension of the spring. The spring is anchored to a very massive support and the block rests on a frictionless surface. After the bullet embeds itself in the block, the block and bullet move as one object compressing the spring by 12 cm before coming -- momentarily - to rest. You may neglect the mass of the spring. Compute the speed of the bullet just before impact BONUS (3 Points) Compute the quantity of energy dissipated by non-conservative forces during the collision between block and bullet.

Explanation / Answer

mass of bullet : m = 15 gram=.015 kg

mass of block: M = 985 gram=0.985 kg

spring's constant :k =1000 N/mtr

displacement :x=12 cm=0.12mtr

speed before collision :u =?

common speed of bullet and block after collision : v

linear momentum before and after the colision of bullet will be same:

mu= (M+m)v

v=mu/M+m..... (1)

after collision spring will keep on compressing till the whole kinetic energy gets stored in form of spring's potential energy

1/2(m+M)v2 =1/2(kx2) =>putting the above values

v = 3.79 m/s ; by eq (1)

u=3.79*1000/15=253 m/s ...... ans

energy lost during collision = initial Kinetic energy- energy stored in spring

=1/2mu2-1/2kx2=480-7.2

=478.2 Joule

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