A 10kg block slides towards the 3kg moving block to create a totally inelastic head on collision. The 3kg block comes to a stop after the collision.
1. What speed must the 10kg block be moving at prior to the collision for this to occur?
2. Instead of sliding along a frictionless surface, suppose the 3kg block is now dropped from rest. After free falling through a height of 1m, it comes into contact with a vertically aligned massless spring. The block compresses the spring by 10m from its equilibrium position before finally coming to a stop. What is the spring constant k?
To answer the given questions, we need to apply the principles of conservation of momentum and conservation of mechanical energy.
1. To determine the initial speed of the 10kg block, we can use the concept of conservation of momentum. According to this principle, the total momentum before the collision must be equal to the total momentum after the collision.
In this case, the momentum of the 10kg block before the collision is given by the product of its mass and its velocity. Let's denote the initial velocity of the 10kg block as v:
Momentum before the collision = Mass of 10kg block * Initial velocity of 10kg block = 10kg * v
After the collision, the 10kg and 3kg blocks stick together and move as a single object. Since the 3kg block comes to rest, the final velocity of the combined 10kg and 3kg blocks will also be 0.
Momentum after the collision = (Mass of 10kg block + Mass of 3kg block) * Final velocity of combined blocks = (10kg + 3kg) * 0 = 0
According to the principle of conservation of momentum, the total momentum before the collision must be equal to the total momentum after the collision:
10kg * v = 0
Simplifying the equation, we find:
v = 0
Therefore, the initial speed of the 10kg block must be 0 prior to the collision.
2. To determine the spring constant k, we can use the concept of conservation of mechanical energy. According to this principle, the initial mechanical energy of the system (before the block comes into contact with the spring) must be equal to the final mechanical energy of the system (after the block comes to a stop).
The initial mechanical energy of the system is given by the potential energy of the block when it is dropped from rest at a height of 1m:
Initial mechanical energy = m * g * h
where m is the mass of the 3kg block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of 1m.
The final mechanical energy of the system is given by the potential energy stored in the compressed spring:
Final mechanical energy = (1/2) * k * x^2
where k is the spring constant and x is the compression of the spring, which is 10m in this case.
According to the principle of conservation of mechanical energy, the initial mechanical energy must be equal to the final mechanical energy:
m * g * h = (1/2) * k * x^2
Substituting the given values:
3kg * 9.8 m/s^2 * 1m = (1/2) * k * (10m)^2
Simplifying the equation, we find:
29.4 kg m^2/s^2 = 50k
Dividing both sides of the equation by 50, we find:
0.588 kg m^2/s^2 = k
Therefore, the spring constant k is approximately 0.588 kg m^2/s^2.