A light, ideal spring with a spring constant k = 100 N/m and uncompressed length L = 0.30 m is mounted to the fixed end of a frictionless plane inclined at an angle θ = 30.0° as shown above. Then a mass M 0.300 kg is affixed to the end of the spring and allowed to slowly slide down the plane until it is resting in an equilibrium position.
A second mass (m = 0.200 kg) is placed at a distance of 0.30 m above the equilibrium position (xo) of the mass-spring system. The second mass (m) is released so that it slides down the ramp and collides with the first mass (M).
a. Calculate the speed of mass, m just before it strikes mass, M.
b. Determine the speed of mass, M and m just after they stick together in a perfectly inelastic collision.
a. To calculate the speed of mass m just before it strikes mass M, we can first find the gravitational potential energy of mass m when it is at a height 0.30 m above the equilibrium position. This is given by:
PE_gravity = m * g * h
where m = 0.200 kg, g = 9.81 m/s², and h = 0.30 m.
PE_gravity = 0.200 * 9.81 * 0.30 = 0.5886 J
Since the ramp is frictionless, all of this potential energy gets converted into kinetic energy of mass m just before it strikes mass M. The kinetic energy of mass m at this point can be written as:
KE_m = 0.5 * m * v^2
where v is the speed of mass m just before it strikes mass M. From conservation of energy, we can write:
PE_gravity = KE_m
Substituting the expressions for PE_gravity and KE_m, we get:
0.5886 = 0.5 * 0.200 * v^2
Now we can solve this equation for v:
v^2 = (0.5886 / (0.5 * 0.200))
v^2 = 5.886
v = √5.886
v ≈ 2.426 m/s
So the speed of mass m just before it strikes mass M is approximately 2.426 m/s.
b. Now we can determine the speed of mass M and m just after they stick together in a perfectly inelastic collision. To do this, we will use conservation of linear momentum. The total momentum before the collision can be written as the sum of the individual momenta of mass M and mass m:
M_i = M*v_M + m*v_m
where M_i is the initial total momentum, v_M is the speed of mass M just before the collision (which is 0), and v_m is the speed of mass m just before the collision. The total momentum after the collision can be written as:
M_f = (M + m)*v_f
where M_f is the final total momentum and v_f is the speed of mass M and m just after they stick together in a perfectly inelastic collision. Since momentum is conserved, we have:
M_i = M_f
Substituting the expressions for M_i and M_f, we get:
M * 0 + m * v_m = (M + m) * v_f
Now we can solve this equation for v_f:
v_f = m * v_m / (M + m)
v_f = 0.200 * 2.426 / (0.300 + 0.200)
v_f ≈ 0.9704 m/s
So the speed of mass M and m just after they stick together in a perfectly inelastic collision is approximately 0.9704 m/s.
To calculate the speed of the second mass, m, just before it strikes the first mass, M, we need to use the principle of conservation of mechanical energy.
a. First, let's find the potential energy of the second mass, m, when it is at a distance of 0.30 m above the equilibrium position, xo. The potential energy, PE, is given by the equation:
PE = m * g * h
Where m is the mass of the second mass (0.200 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height above the equilibrium position (0.30 m).
PE = 0.200 kg * 9.8 m/s² * 0.30 m
PE = 0.588 J
b. Now, let's find the potential energy of the second mass, m, when it reaches the equilibrium position, xo. At this point, all the potential energy is converted into kinetic energy, KE. Therefore, the kinetic energy of mass, m, just before it strikes mass, M, is equal to the potential energy at the equilibrium position.
KE = PE = 0.588 J
c. Using the equation for kinetic energy, KE = 0.5 * m * v², we can solve for v, the velocity of mass m just before it strikes mass M.
0.5 * m * v² = 0.588 J
0.5 * 0.200 kg * v² = 0.588 J
v² = (0.588 J) / (0.5 * 0.200 kg)
v² = 5.88 m²/s²
Taking the square root of both sides, we find:
v = √(5.88 m²/s²)
v ≈ 2.42 m/s
The speed of mass m just before it strikes mass M is approximately 2.42 m/s.
b. After the collision, mass m and M stick together and move as a single combined mass. To determine their speed just after the collision, we can apply the principle of conservation of momentum.
Since the collision is perfectly inelastic, the two masses stick together and move with a common final velocity, V.
The momentum before the collision is given by:
momentum_before = m * v_m + M * v_M
Where m is the mass of the second mass (0.200 kg), v_m is the velocity of mass m just before the collision (2.42 m/s), M is the mass of the first mass (0.300 kg), and v_M is the velocity of mass M just before the collision (which we need to find).
The momentum after the collision is given by:
momentum_after = (m + M) * V
Applying the principle of conservation of momentum, we set the momentum before the collision equal to the momentum after the collision:
momentum_before = momentum_after
m * v_m + M * v_M = (m + M) * V
Substituting the known values:
0.200 kg * 2.42 m/s + 0.300 kg * v_M = (0.200 kg + 0.300 kg) * V
Simplifying the equation:
0.484 kg + 0.300 kg * v_M = 0.500 kg * V
Now, solving for V:
0.300 kg * v_M = 0.500 kg * V - 0.484 kg
0.300 kg * v_M = 0.500 kg * V - 0.484 kg
v_M = (0.500 kg * V - 0.484 kg) / (0.300 kg)
Now, substitute this value of v_M into the equation for momentum before the collision in order to find V:
0.200 kg * 2.42 m/s + 0.300 kg * [(0.500 kg * V - 0.484 kg) / (0.300 kg)] = (0.200 kg + 0.300 kg) * V
Simplifying the equation:
0.484 kg + 0.300 kg * [(0.500 kg * V - 0.484 kg) / (0.300 kg)] = 0.500 kg * V
Now, solve for V.
To solve this problem, we will break it down into three parts:
1. Find the equilibrium position (xo) and potential energy (PE) of the mass-spring system.
2. Calculate the speed of mass m just before it strikes mass M.
3. Determine the speed of mass M and m just after they stick together in a perfectly inelastic collision.
Let's tackle each part step-by-step.
Step 1: Find the equilibrium position (xo) and potential energy (PE)
Given:
- Spring constant (k) = 100 N/m
- Uncompressed length of the spring (L) = 0.30 m
- Inclined plane angle (θ) = 30.0°
- Mass of M (M) = 0.300 kg
To find the equilibrium position (xo), we need to determine the total force acting on the mass M at equilibrium. The forces acting on M are its weight (mg) acting straight downwards and the spring force (kΔx) acting upwards along the inclined plane.
At equilibrium, these forces balance each other:
mg = kΔx
Solving for Δx:
Δx = mg / k
Plugging in the given values:
Δx = (0.300 kg * 9.8 m/s^2) / (100 N/m)
= 0.0294 m
Therefore, the equilibrium position (xo) of the mass-spring system is 0.0294 m.
Next, let's calculate the potential energy (PE) of the system at equilibrium:
PE = 0.5 * k * Δx^2
PE = 0.5 * 100 N/m * (0.0294 m)^2
= 0.043 N
Therefore, the potential energy (PE) of the mass-spring system at equilibrium is 0.043 N.
Step 2: Calculate the speed of mass m just before it strikes mass M.
Given:
- Mass of m (m) = 0.200 kg
- The distance above the equilibrium position (xo) = 0.30 m
The potential energy (PE) of mass m at the top position is converted into kinetic energy as it slides down the ramp. Therefore, we can calculate the speed just before it strikes mass M using the conservation of energy:
PE (at top) = KE (at impact)
PE (at top) = m * g * h
KE (at impact) = 0.5 * m * v^2
Plugging in the given values:
m * g * h = 0.5 * m * v^2
(0.200 kg) * (9.8 m/s^2) * (0.30 m) = 0.5 * (0.200 kg) * v^2
v^2 = (2 * (0.200 kg) * (9.8 m/s^2) * (0.30 m)) / (0.200 kg)
v^2 = 5.88 m^2/s^2
v = √(5.88 m^2/s^2)
v ≈ 2.42 m/s
Therefore, the speed of mass m just before it strikes mass M is approximately 2.42 m/s.
Step 3: Determine the speed of mass M and m just after they stick together in a perfectly inelastic collision.
During an inelastic collision, the two masses stick together and move with a common final velocity. To find this velocity, we can apply the conservation of momentum:
Initial momentum = Final momentum
(m * vm_initial) + (M * vM_initial) = (m + M) * vf
Where:
- vm_initial = initial velocity of mass m
- vM_initial = initial velocity of mass M
- vf = final velocity of the combined masses
We know that mass m has an initial velocity of 2.42 m/s (as calculated in Step 2) and mass M is initially at rest (vM_initial = 0 m/s).
Plugging in the given values:
(0.200 kg * 2.42 m/s) + (0.300 kg * 0 m/s) = (0.200 kg + 0.300 kg) * vf
0.484 kg·m/s = 0.500 kg * vf
vf = 0.484 kg·m/s / 0.500 kg
vf = 0.968 m/s
Therefore, the velocity of the combined masses just after the collision is approximately 0.968 m/s.
To summarize the answers to the given questions:
a. The speed of mass m just before it strikes mass M is approximately 2.42 m/s.
b. The speed of mass M and m just after they stick together in a perfectly inelastic collision is approximately 0.968 m/s.