Question

An 8.0-g bullet is fired into a 4.0-kg block that is initially at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into an ideal massless spring and compresses it by 8.5 cm. The spring constant k The spring constant of he spring is 2400 N/m. The initial velocity of the bullet is closest to___________.

Answer 1

vi = 1042.08 m/s

Step-by-step explanation:

Given:-

- The mass of the bullet, mg = 8.0 g

- The mass of the block, mb = 4.0 kg

- The initial seed of the bullet = vi

- The block was initially at rest.

- The spring with stiffness constant, k = 2400 N/m

- The compression of spring, x = 8.5 cm

Find:-

The initial velocity of the bullet is closest to

Solution:-

- We will consider the system of block and bullet embedded inside has a kinetic energy ( K.E ). Since there are no friction forces acting on the block and no work is being done on the block while it moves, then the system is taken to be in isolation and principle of conservation of energy is valid.

- The kinetic energy (K.E) of bullet embedded block is stored as elastic potential energy (E.P) in the spring when it is compressed.

K.E = E.P

0.5*(mb+mg)*vf^2 = 0.5*k*x^2

Where, vf is the velocity of bullet plus block after impact.

vf = √(k/mb+mg)*x

vf = √(2400/4 + 0.008)*0.085

vf = 2.08 m/s

- The velocity after impact is vf = 2.08 m/s

- The principle of conservation of linear momentum is also valid for the same system.

Pi = Pf

Where,

Pi : initial momentum of system ( before impact)

Pf: Final momentum of system (after impact)

mg*vi = ( mg + mb )*vf

vi = ( 1 + mb/mg )*vf

vi = ( 1 + 4/0.008 )*2.08

vi = 1042.08 m/s

Answer 2

12,750m/s

Step-by-step explanation:

The change of momentum of the bodies = Force that causes the spring to extend

momentum = mass × velocity

Force on the spring = 1/2ke²

K is the spring constant

e is the extension

Since momentum = mass m × velocity v

The equation becomes

m1u1+m2u2 = 1/2ke²

m1 and u1 are the masses of the bullet and block respectively.

u1 and u2 are the velocities of bullet and block respectively

Given m1 = 8.0g = 0.008kg

u1= ?

m2 =4kg

u2 = 0m/s (block initially at rest)

Substituting this values in the formula

0.008u1+4(0) = 1/2× 2400 × 0.085

0.008u1 =102

u1 = 102/0.008

u1 = 12,750m/s