A block of mass m1 = 1.00 kg slides along a frictionless table with a speed of 16.0 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 6.00 kg moving at 8.00 m/s. A massless spring with a spring constant k = 1320 N/m is attached to the backside of m2. When the blocks collide, what is the maximum compression (in meters) of the spring?
solved
To find the maximum compression of the spring during the collision, we need to use the principle of conservation of momentum and conservation of energy. Here's how to approach the problem:
1. Determine the initial momentum of both blocks before the collision. Momentum (p) is given by the product of mass (m) and velocity (v), so the momentum (p1) of block m1 is given by p1 = m1 * v1, and the momentum (p2) of block m2 is given by p2 = m2 * v2.
p1 = m1 * v1
= 1.00 kg * 16.0 m/s
= 16.0 kg m/s
p2 = m2 * v2
= 6.00 kg * 8.00 m/s
= 48.0 kg m/s
2. Apply the principle of conservation of momentum. In an isolated system, the total momentum before a collision equals the total momentum after the collision. Therefore, p1 + p2 = p1' + p2', where p1' and p2' represent the final momenta of the blocks.
p1 + p2 = p1' + p2'
3. In this case, after the collision, the two blocks will stick together and move as one. Thus, the final velocity (vf) of the system can be found by dividing the total momentum by the total mass (m1 + m2).
vf = (p1 + p2) / (m1 + m2)
= (16.0 kg m/s + 48.0 kg m/s) / (1.00 kg + 6.00 kg)
= 64.0 kg m/s / 7.00 kg
≈ 9.143 m/s
4. Now, let's find the compression of the spring. At maximum compression, all of the initial kinetic energy of the blocks will be converted into potential energy stored in the spring. We can use the equation for potential energy of a spring, U = (1/2) * k * x^2, where U is the potential energy, k is the spring constant, and x is the compression of the spring.
Initial Kinetic Energy = Potential Energy at maximum compression
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * k * x^2
Substitute the given values:
(1/2) * 1.00 kg * (16.0 m/s)^2 + (1/2) * 6.00 kg * (8.00 m/s)^2 = (1/2) * 1320 N/m * x^2
Simplify and solve for x:
128 J + 192 J = 660 N/m * x^2
320 J = 660 N/m * x^2
x^2 = (320 J) / (660 N/m)
x^2 ≈ 0.485 m^2
Take the square root of both sides to find x:
x ≈ √(0.485 m^2)
x ≈ 0.697 m
So, the maximum compression of the spring during the collision is approximately 0.697 meters.