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Question
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?
1. To determine the speed of the 10kg block prior to the collision, we can use the law of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's assume the initial velocity of the 10kg block is v1 and the initial velocity of the 3kg block is v2. The final velocity of both blocks after the collision is 0 since the 3kg block comes to a stop.
The momentum before the collision is given by:
Initial momentum of the 10kg block = mass of the 10kg block * velocity of the 10kg block
= 10kg * v1
Initial momentum of the 3kg block = mass of the 3kg block * velocity of the 3kg block
= 3kg * v2
According to the conservation of momentum:
Initial momentum of the 10kg block + Initial momentum of the 3kg block = Final momentum after the collision
10kg * v1 + 3kg * v2 = 0
Since the 3kg block comes to a stop after the collision, v2 = 0:
10kg * v1 + 3kg * 0 = 0
10kg * v1 = 0
v1 = 0
Therefore, the 10kg block must be at rest prior to the collision in order for the 3kg block to come to a stop after the collision.
2. To determine the spring constant k, we can use the principles of potential energy.
The potential energy of the 3kg block when it is dropped from a height of 1m is given by:
Potential energy = mass * gravitational acceleration * height
= 3kg * 9.8 m/s^2 * 1m
= 29.4 J
The potential energy is converted into the potential energy stored in the compressed spring when the block comes to a stop. The energy stored in a spring is given by:
Potential energy stored in the spring = (1/2) * k * x^2
where k is the spring constant and x is the compression distance.
Plugging in the given values:
29.4 J = (1/2) * k * (10m)^2
29.4 J = 50k
k = 29.4 J / 50
k = 0.588 N/m
Therefore, the spring constant k is 0.588 N/m.
To answer these questions, we need to apply the principles of Newton's laws of motion and conservation of energy.
1. What speed must the 10kg block be moving at prior to the collision for this to occur?
To solve this question, we can use the principle of conservation of momentum. In an inelastic collision, the total momentum before and after the collision remains the same. The momentum of an object is given by the product of its mass and velocity.
Let's denote the velocity of the 10kg block before the collision as v1 and the velocity of the 3kg block after the collision as v2. Since the collision is totally inelastic, both blocks will have the same final velocity.
Conservation of momentum equation:
(m1 × v1) + (m2 × 0) = (m1 + m2) × v2
In this case, m1 = 10kg and m2 = 3kg.
With these values plugged into the equation above, we get:
(10kg × v1) + (3kg × 0) = (10kg + 3kg) × v2
10kg × v1 = 13kg × v2
To find the relationship between v1 and v2, we divide both sides of the equation by 10:
v1 = (13/10) × v2
Since both blocks have the same final velocity v2, we can substitute v2 with v:
v1 = (13/10) × v
This means that the speed of the 10kg block before the collision must be 13/10 times the final velocity of the two blocks.
2. What is the spring constant k?
To solve this question, we can use the principle of conservation of mechanical energy. In this case, the potential energy of the block when it reaches its maximum height after free falling is converted into the potential energy stored in the compressed spring.
The potential energy of an object is given by the product of its mass, acceleration due to gravity, and height. In this case, the mass of the block is 3kg, and the height is 10m.
Potential energy equation:
Potential energy (PE) = mass × gravity × height
PE = 3kg × 9.8m/s^2 × 10m
PE = 294 J (joules)
The potential energy at the maximum height is equal to the potential energy stored in the compressed spring. This energy can be calculated using Hooke's Law, which states that the potential energy stored in a spring is proportional to the square of its displacement (x) from its equilibrium position (10m in this case) and its spring constant (k).
Potential energy equation (using Hooke's Law):
PE = (1/2) × k × x^2
We know that the potential energy is 294 joules and the displacement (x) is 10m. Plugging in these values into the equation above, we get:
294 J = (1/2) × k × 10m^2
294 J = 5k
Solving for k, we divide both sides by 5:
k = 294 J / 5
k = 58.8 N/m
Therefore, the spring constant (k) is 58.8 N/m.