A mass 1.2 kg of is attached to a spring and placed on a friction-less air track. Initially the mass is resting in equilibrium. A 10-gram bullet is fired into the mass to initiate oscillatory motion. The bullet hits the mass from the left side at a speed of 120 m/s at t=0 and gets lodged inside of it. After that, the system starts oscillating harmonically with a period of 15 seconds.
1. Use momentum conservation to find the initial velocity of the mass right after it was hit by the
bullet.
2. What was the kinetic energy of the bullet? What was the kinetic energy of the mass right after it
was hit by the bullet? Where did the difference go?
3. Find the amplitude and phase constant of the resulting oscillations.
4. Write explicit expressions for the position x(t) and velocity v(t) of the mass as a function of time.
To answer these questions, we will need to apply different principles of physics. Let's go through each question step by step:
1. Use momentum conservation to find the initial velocity of the mass right after it was hit by the bullet:
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the bullet is fired is zero since there is no initial motion. After the bullet is fired and lodges in the mass, the system starts moving.
The momentum before the collision (bullet) is given by:
Initial momentum of bullet = mass of bullet * velocity of bullet
= (0.01 kg) * (120 m/s) = 1.2 kg⋅m/s
Considering the momentum conservation, the momentum of the bullet and the mass after the collision would be equal to the initial momentum of the system, which is zero.
Hence,
0 = (mass of bullet + mass of mass) * final velocity
(0.01 kg + 1.2 kg) * final velocity = 0
final velocity = 0 m/s
Therefore, the initial velocity of the mass right after it was hit by the bullet is 0 m/s.
2. What was the kinetic energy of the bullet? What was the kinetic energy of the mass right after it was hit by the bullet? Where did the difference go?
The kinetic energy of an object is given by the equation:
Kinetic energy (KE) = (1/2) * mass * velocity^2
For the bullet:
KE of bullet = (1/2) * mass of bullet * (velocity of bullet)^2
= (1/2) * 0.01 kg * (120 m/s)^2 = 72 J
For the mass right after the collision:
KE of mass = (1/2) * mass of mass * (final velocity)^2
= (1/2) * 1.2 kg * (0 m/s)^2 = 0 J
The difference in kinetic energy (72 J - 0 J = 72 J) went into potential energy stored in the spring during the collision. This energy is converted into potential energy rather than kinetic energy because the system started oscillating harmonically, implying that the motion is oscillatory and energy is being exchanged between kinetic and potential forms.
3. Find the amplitude and phase constant of the resulting oscillations:
To find the amplitude and phase constant of the resulting oscillations, we need the period (T) and the initial conditions (position and velocity) of the mass.
Given that the system starts oscillating harmonically with a period of 15 seconds, we have:
Period (T) = 15 s
The amplitude (A) can be calculated using the formula:
Amplitude (A) = displacement (x) max = (initial position (x0) - equilibrium position) = 0 - 0 = 0
Since the initial position is the equilibrium position, the amplitude is zero.
The phase constant (φ) is determined by the initial conditions (position and velocity) of the mass. However, this information is not provided in the given problem statement. Without the initial conditions, we cannot determine the phase constant.
4. Write explicit expressions for the position x(t) and velocity v(t) of the mass as a function of time:
The general form of the position (x(t)) and velocity (v(t)) equations for a mass-spring system undergoing simple harmonic motion are:
x(t) = A * cos(ωt + φ) (Equation 1)
v(t) = -A * ω * sin(ωt + φ) (Equation 2)
Where:
- A is the amplitude (already determined as 0 in this case)
- ω is the angular frequency, given by ω = (2π / T), where T is the period
- φ is the phase constant (unknown in this case)
- t is the time
Since the amplitude is zero, the position equation (Equation 1) simplifies to:
x(t) = 0
And the velocity equation (Equation 2) simplifies to:
v(t) = 0
Therefore, the position and velocity of the mass as a function of time are both zero.