Problem 2- A 15 kg block is attached to a very light horizontal spring of force constant 400 N/m and is resting on a smooth horizontal table as shown in the figure below. Suddenly it is struck by a 3 kg stone traveling horizontally at 8 m/s to the right, whereupon the stone rebounds at 2 m/s horizontally to the left.
a) Find the speed of the 15 kg mass just after the collision.
b) Is the collision elastic?
c) Find the maximum distance that the block will compress the spring after the collision.
a) After the collision, the total momentum is conserved. Initially, only the stone has momentum, so its initial momentum is given by the equation:
momentum = mass × velocity
momentum = 3 kg × 8 m/s
momentum = 24 kg·m/s
After the collision, both the stone and the block have momentum. Let's assume the speed of the block just after the collision is v m/s. The momentum of the stone and block after the collision is given by:
momentum = (3 kg × 2 m/s) + (15 kg × v m/s)
24 kg·m/s = 6 kg·m/s + 15 kg·v
Simplifying the equation, we get:
24 kg·m/s - 6 kg·m/s = 15 kg·v
18 kg·m/s = 15 kg·v
v = 18 kg·m/s ÷ 15 kg
v ≈ 1.2 m/s
So, the speed of the 15 kg mass just after the collision is approximately 1.2 m/s.
b) To determine if the collision is elastic, we need to compare the total kinetic energy before the collision with the total kinetic energy after the collision.
The total kinetic energy before the collision is given by:
KE_before = (0.5) × mass_stone × velocity_stone^2
KE_before = (0.5) × 3 kg × (8 m/s)^2
KE_before = 0.5 × 3 kg × 64 m^2/s^2
KE_before = 96 J
The total kinetic energy after the collision is given by:
KE_after = (0.5) × mass_stone × velocity_stone^2 + (0.5) × mass_block × velocity_block^2
KE_after = (0.5) × 3 kg × (2 m/s)^2 + (0.5) × 15 kg × (1.2 m/s)^2
KE_after = 0.5 × 3 kg × 4 m^2/s^2 + 0.5 × 15 kg × 1.44 m^2/s^2
KE_after = 6 J + 10.8 J
KE_after = 16.8 J
Since KE_after ≠ KE_before, the total kinetic energy after the collision is not equal to the total kinetic energy before the collision. Therefore, the collision is not elastic.
c) To find the maximum compression of the spring, we can use conservation of mechanical energy. Initially, the only form of mechanical energy is potential energy due to the spring, which is zero. After compressing the spring, the maximum potential energy is given by:
potential energy = (0.5) × force_constant × maximum_distance^2
Equating the maximum potential energy to the initial kinetic energy (KE_before), we can find the maximum distance (x) that the block will compress the spring:
(0.5) × force_constant × maximum_distance^2 = KE_before
(0.5) × 400 N/m × maximum_distance^2 = 96 J
200 N/m × maximum_distance^2 = 96 J
maximum_distance^2 = 96 J / 200 N/m
maximum_distance^2 = 0.48 m
maximum_distance = √(0.48 m)
maximum_distance ≈ 0.69 m
Therefore, the maximum distance that the block will compress the spring after the collision is approximately 0.69 meters.
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of mechanical energy.
a) Find the speed of the 15 kg mass just after the collision:
We can start by calculating the initial momentum of the system (before the collision) and the final momentum of the system (after the collision).
Initial momentum (before the collision):
The initial momentum of the system is the sum of the momenta of the block and the stone:
P_initial = m_block * v_block + m_stone * v_stone
= (15 kg) * 0 + (3 kg) * (8 m/s)
= 24 kg*m/s
Final momentum (after the collision):
The final momentum after the collision can be determined by considering that the 15 kg block and the 3 kg stone will move together as one system:
P_final = (15 kg + 3 kg) * v_final
According to the principle of conservation of momentum:
P_initial = P_final
24 kg*m/s = (18 kg) * v_final
v_final = 24 kg*m/s / 18 kg
v_final ≈ 1.33 m/s
Therefore, the speed of the 15 kg mass just after the collision is approximately 1.33 m/s.
b) Is the collision elastic?
To determine if the collision is elastic, we need to check if the kinetic energy is conserved.
Initial kinetic energy (before the collision):
K_initial = 1/2 * m_block * v_block^2 + 1/2 * m_stone * v_stone^2
= 1/2 * (15 kg) * 0^2 + 1/2 * (3 kg) * (8 m/s)^2
= 96 J
Final kinetic energy (after the collision):
K_final = 1/2 * (15 kg + 3 kg) * v_final^2
= 1/2 * (18 kg) * (1.33 m/s)^2
≈ 14.97 J
The kinetic energy is not conserved since K_final ≠ K_initial, so the collision is not elastic.
c) Find the maximum distance that the block will compress the spring after the collision:
To find the maximum distance that the block will compress the spring, we can use the principle of conservation of mechanical energy.
The maximum compression of the spring occurs when the entire system (block and spring) reaches potential energy maximum.
Potential energy of the spring:
U_spring = 1/2 * k * x^2
where k is the force constant of the spring and x is the compression distance.
Since the kinetic energy is not conserved, we can only consider the change in total mechanical energy.
ΔE = ΔK + ΔU
Since the initial kinetic energy is 96 J and the final kinetic energy is approximately 14.97 J, we have:
ΔE = 96 J - 14.97 J
ΔE ≈ 81.03 J
The change in potential energy of the spring is equal to the change in total mechanical energy since there is no change in gravitational potential energy.
ΔU = ΔE = 81.03 J
We can set the change in potential energy equal to 81.03 J and solve for the maximum compression distance (x):
ΔU = 1/2 * k * x^2
81.03 J = 1/2 * (400 N/m) * x^2
162.06 J/(N/m) = x^2
x ≈ √(162.06 J/(N/m))
x ≈ 12.75 m
Therefore, the maximum distance that the block will compress the spring after the collision is approximately 12.75 cm.
To solve this problem, we can apply the conservation of linear momentum and the conservation of mechanical energy principles. Let's break down each part of the problem step by step:
a) Find the speed of the 15 kg mass just after the collision:
The conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision. The momentum (p) can be calculated as the product of mass (m) and velocity (v).
The initial momentum before the collision is given by:
p_initial = (mass of stone) * (initial velocity of stone)
= (3 kg) * (8 m/s)
The final momentum after the collision is given by:
p_final = (mass of stone) * (final velocity of stone) + (mass of block) * (final velocity of the block)
Since the initial and final velocities of the block are not given, we need to find them using the given information. We know that the stone rebounds at 2 m/s horizontally to the left after striking the block. Therefore, the final velocity of the block just after the collision is also 2 m/s horizontally to the left.
Substituting these values in the conservation of linear momentum equation, we have:
p_initial = p_final
(3 kg) * (8 m/s) = (3 kg) * (2 m/s) + (15 kg) * (final velocity of the block)
Simplifying the equation, we can solve for the final velocity of the block.
b) Is the collision elastic?
To determine whether the collision is elastic or not, we need to analyze the conservation of mechanical energy. In an elastic collision, both linear momentum and mechanical energy are conserved. If the mechanical energy is conserved, the collision is elastic. Otherwise, it is inelastic.
The mechanical energy (E) can be calculated as the sum of kinetic energies of the stone and the block.
The initial mechanical energy before the collision is given by:
E_initial = (1/2) * (mass of stone) * (initial velocity of stone)^2
The final mechanical energy after the collision is given by:
E_final = (1/2) * (mass of stone) * (final velocity of stone)^2 + (1/2) * (mass of block) * (final velocity of the block)^2
If E_initial = E_final, the collision is elastic. Otherwise, it is inelastic.
c) Find the maximum distance that the block will compress the spring after the collision:
To find the maximum distance that the block will compress the spring after the collision, we need to analyze the potential energy stored in the spring. The potential energy (U) can be calculated using Hooke's Law:
U = (1/2) * (force constant of the spring) * (compression distance)^2
Since the block will compress the spring, the potential energy stored in the spring will be equal to the change in mechanical energy of the block after the collision:
Change in mechanical energy = E_final - E_initial
Substituting the values and solving for the compression distance, we can find the maximum distance that the block will compress the spring.
Remember to convert units as necessary and ensure that the signs of velocities and distances are consistent with their respective directions.
Plus is right
original momentum = 3*8 = 24
so final momentum = 24
What does right after the collision mean I wonder. Nothing happens when it first hits the spring. I will look at the final outcome.
24 = -3(2) + 15 v
24 = -6 + 15 v
30 = 15 v
v = 2 m/s part a
Initially
(1/2) m v^2 = .5*3 * 64 = 96 Joules
Finally
.5*3*4 + .5*15*4 = 36 Joules
Not elastic part b
Oh my we lost some energy in that spring
I will say all that lost energy went into the spring, half on compression and half on expansion
(96-36)/2 = 30 joules = .5 k x^2
.5 * 400* x^2 = 30
x = .387 m