Respuesta :
Answer:
12,750m/s
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
Answer:
vi = 1042.08 m/s
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
