Answer:
The distance the block will slide before it stops is x= 3.4 m .
Explanation:
m1= 0,02 kg
V1= 400 m/s
m2= 2 kg
V2= ?
g= 9.8 m/s²
μ= 0.24
N = m2 * g = 19.6 N
Fr= μ * N
Fr= 4.704 N
due to the conservation of the amount of movement:
m1*V1 = m2*V2
V2= 4 m/s
Fr = m2*a
a= Fr/m2
a= -2.352 m/s²
matching momentum and amount of movement
Fr*t=m2*V2
t= 1.7 sec
x= V2*t - (a*t²)/2
x= 3.4 m
Answer:
The distance the block will slide before it stops is 3.3343 m
Explanation:
Given;
mass of bullet, m₁ = 20-g = 0.02 kg
speed of the bullet, u₁ = 400 m/s
mass of block, m₂ = 2-kg
coefficient of kinetic friction, μk = 0.24
Step 1:
Determine the speed of the bullet-block system:
From the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the bullet-block system after collision
(0.02 x 400) + (2 x 0) = v (0.02 + 2)
8 = v (2.02)
v = 8/2.02
v = 3.9604 m/s
Step 2:
Determine the time required for the bullet-block system to stop
Apply the principle of conservation momentum of the system
[tex]v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f[/tex]
when the system stops, vf = 0
3.9604 -2.352t = 0
2.352t = 3.9604
t = 3.9604/2.352
t = 1.684 s
Thus, time required for the system to stop is 1.684 s
Finally, determine the distance the block will slide before it stops
From kinematic, distance is the product of speed and time
[tex]S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2[/tex]
Now, recall that t = 1.684 s
S = 3.9604(1.684) - 1.176(1.684)²
S = 6.6693 - 3.3350
S = 3.3343 m
Thus, the distance the block will slide before it stops is 3.3343 m
