Respuesta :
(a) 0.147
First of all, we apply the law of conservation of momentum to find the velocity of the block when the bullet emerges from it:
[tex]mu +MU = mv +MV[/tex]
where
m = 4.00 g = 0.004 kg is the mass of the bullet
M = 0.800 kg is the mass of the block
u = 360 m/s is the initial velocity of the bullet
U = 0 is the initial velocity of the block
v = 120 m/s is the final velocity of the bullet
V = ? is the velocity of the block after the bullet emerges from it
Solving for V,
[tex]V=\frac{mu-mv}{M}=\frac{(0.004)(360)-(0.004)(120)}{0.8}=1.2 m/s[/tex]
Now we consider the second part, in which the block slides after the bullet emerges from it. We now that the block slides a distance of
s = 50.0 cm = 0.50 m
So, we can find its acceleration by using the suvat equation:
[tex]v^2-u^2=2as[/tex]
where the new notation is:
v = 0 is the final velocity of the block
u = 1.2 m/s is the initial velocity of the block
a is the acceleration
s = 0.50 m is the stopping distance
Solving for a,
[tex]a=\frac{v^2-u^2}{2s}=\frac{0-1.2^2}{2(0.50)}=-1.44 m/s^2[/tex]
The frictional acting between the block and the surface is
[tex]F=-\mu mg[/tex]
where
[tex]\mu[/tex] is the coefficient of kinetic friction
m is the mass of the block
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
And using Newton's second law [tex]F=ma[/tex], we can find the value of the coefficient of friction:
[tex]ma=-\mu mg\\\mu = -\frac{a}{g}=-\frac{-1.44}{9.8}=0.147[/tex]
(b) -230.4 J
The initial kinetic energy of the bullet is
[tex]K_i=\frac{1}{2}mu^2[/tex]
where
m = 4.00 g = 0.004 kg is the mass of the bullet
u = 360 m/s is the initial velocity of the bullet
Substituting,
[tex]K_i = \frac{1}{2}(0.004)(360)^2=259.2 J[/tex]
While its final kinetic energy is
[tex]K_f=\frac{1}{2}mv^2[/tex]
where
v = 120 m/s is the final velocity of the bullet
Substituting,
[tex]K_f = \frac{1}{2}(0.004)(120)^2=28.2 J[/tex]
So, the decrease in kinetic energy of the bullet is
[tex]\Delta K = K_f - K_i = 28.2-259.2=-230.4 J[/tex]
(c) 0.58 J
The kinetic energy of the block at the instant the bullet leaves the block is
[tex]K=\frac{1}{2}MV^2[/tex]
where
M = 0.800 kg is the mass of the block
V = 1.2 m/s is the velocity of the block (found in part a)
Substituting into the equation, we find
[tex]K=\frac{1}{2}(0.800)(1.2)^2=0.58 J[/tex]
Note that the kinetic energy gained by the block is not equal to the kinetic energy lost by the bullet: this is because the collision is not elastic, as some of the energy is lost due to frictions between the bullet and the block as the bullet passes through the block.
(a) The coefficient of kinetic friction between block and surface is 0.15.
(b) The decrease in kinetic energy of the bullet is 230.4 J.
(c) The kinetic energy of the block at the instant after the bullet passes through it is 0.58 J.
Conservation of linear momentum
The final speed of the block can be determined by applying the principle of conservation of linear momentum as follows;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.004(360) + 0.8(0) = 0.004(120) + 0.8(v₂)
1.44 = 0.48 + 0.8v₂
0.8v₂ = 0.96
v₂ = 0.96/0.8
v₂ = 1.2 m/s
Acceleration of the block is calculated as follows;
v² = u² + 2as
v² = 2as
a = v²/2s
a = (1.2)²/(2 x 0.5)
a = 1.44 m/s²
Coefficient of kinetic friction between block and surface
ma = μmg
a = μg
μ = a/g
μ = 1.44/9.8
μ = 0.15
Decrease in kinetic energy of the bullet
[tex]\Delta K.E = K.E_f - K.E_i\\\\\Delta K.E = \frac{1}{2} m(v_f^2 - v_i^2)\\\\\Delta K.E = \frac{1}{2} \times 0.004 \times (120^2 - 360^2)\\\\\Delta K.E = -230.4 \ J[/tex]
Kinetic energy of the block
The kinetic energy of the block at the instant after the bullet passes through it is calculated as follows;
[tex]K.E = \frac{1}{2} \times 0.8 \times 1.2^2\\\\K.E = 0.58 J[/tex]
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