Hello im kinda new here and hoping someone can help me with this energy conservation question (I believe).
A 2.10g- bullet embeds itself in a 1.30kg - block, which is attached to a spring of force constant 770 N/m .
1) If the maximum compression of the spring is 5.60 cm , find the initial speed of the bullet.
2)Find the time for the bullet-block system to come to rest?
Thank you for any assistance
1) The embedding process is inelastic, so you have to use conservation of momentum to get the final velocity after the bullet is embedded and before spring compression starts. Let
V = initial bullet velocity,
v = velocity after it gets embedded
m = bullet mass
M = block mass
Momentum ocnservation tells you that
V*m = v*(m + M)
v = V*m/(m+M) = 1.613*10^-3 V
Spring compression is an elastic process, so the kinetic energy after embeddig = potential energy of thew fully compressed spring.
(1/2) kX^2 = (1/2)(m+M)v^2
= (1/2)m^2*V^2/(m+M)
Solve for V. Make sure X is in units of meters.
(2) The bullet-block system will keep on oscillating, so I think they want the time it takes for the spring to reach maximum compression (before it bounces back). That would be 1/4 of the period of oscillation, or
(1/4)*(2 pi)sqrt[k/(m+M)]
I agree with the previous post except for the last part, it would be (1/4)*(2 pi)sqrt[(m+M)/k] instead.
To solve these questions, you can use the principle of conservation of energy. The initial kinetic energy of the bullet will be equal to the potential energy stored in the spring when it reaches its maximum compression.
1) To find the initial speed of the bullet, you need to equate the initial kinetic energy of the bullet (½mv²) to the potential energy of the spring (½kx²), where m is the mass of the bullet, v is its initial speed, k is the force constant of the spring, and x is the maximum compression.
Based on the given information:
Mass of the bullet, m = 2.10g = 0.00210 kg
Mass of the block, M = 1.30 kg
Force constant of the spring, k = 770 N/m
Maximum compression, x = 5.60 cm = 0.0560 m
Setting the initial kinetic energy equal to the potential energy:
½mv² = ½kx²
Substituting the known values:
½(0.00210 kg)v² = ½(770 N/m)(0.0560 m)²
Simplifying the equation:
0.00105 v² = 0.019536 Nm
Dividing both sides of the equation by 0.00105:
v² = 18.611 N/m
Taking the square root of both sides of the equation:
v ≈ √(18.611 N/m)
Therefore, the initial speed of the bullet is approximately v ≈ 4.316 m/s.
2) To find the time for the bullet-block system to come to rest, you need to use the concept of mechanical energy conservation. When the bullet-block system comes to rest, all of its initial kinetic energy is converted into potential energy in the spring.
The initial kinetic energy of the bullet is given by ½mv². The potential energy stored in the spring is given by ½kx². Equating these two energies:
½mv² = ½kx²
Substituting the known values:
½(0.00210 kg)v² = ½(770 N/m)(0.0560 m)²
Simplifying the equation:
v² = (770 N/m)(0.0560 m)² / (0.00210 kg)
v² ≈ 0.8428571 Nm/kg
Taking the square root of both sides of the equation:
v ≈ √(0.8428571 Nm/kg)
Therefore, the final speed of the bullet-block system when it comes to rest is approximately v ≈ 0.9191 m/s.
To find the time taken for the system to come to rest, you can use the equation:
v = u + at
where u is the initial velocity (in this case, the initial speed of the bullet), a is the acceleration (which is negative because it opposes the initial velocity), and t is the time taken.
Rearranging the equation:
t = (v - u) / a
Substituting the known values:
t = (0.9191 m/s - 4.316 m/s) / (-0.9191 m/s²)
t ≈ (-3.3969 m/s) / (-0.9191 m/s²)
t ≈ 3.695 s
Therefore, the time taken for the bullet-block system to come to rest is approximately t ≈ 3.695 seconds.