1. The spring has 3.25 J of elastic potential energy when it is compressed dista
ID: 1530322 • Letter: 1
Question
1. The spring has 3.25 J of elastic potential energy when it is compressed distance x. Determine the distance x.
2. Determine the velocity of Block A on the table when it leaves contact with the spring.
3. Determine the velocity of the stuck together blocks immediately following the collision, while they are still on the table.
4. What is the total kinetic energy of the blocks just before they hit the floor.
650 N/m 4.0 kg 4.0 kg DHA Note: Figure not drawn to scale. Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass and spring constant 650Nm. The other end of the spring is attached to a wall. The block is pushed toward the wall until the spring has been compressed a distance x, as shown above. Block A is released, then strikes and sticks to Block Bat the edge of the table. The blocks then fall to the ground following the trajectory shown, falling 0.80 m vertically to the floor. Air resistance is negligible.Explanation / Answer
1. In The spring has 3.25 J of elastic potential energy when it is compressed distance x. Determine the distance x.
PS = ½kx²
x = (2PS/k)
x = (2(3.25) / 650)
x = 0.10 m or 10 cm
2. Determine the velocity of Block A on the table when it leaves contact with the spring.
The spring potential energy will be converted to kinetic energy of the mass
KE = PS
½mv² = PS
v = (2PS/m)
v = (2(3.25 / 4.0)
v = 1.27475...
v = 1.3 m/s
3. Determine the velocity of the stuck together blocks immediately following the collision.
Conservation of momentum
We are not told the mass of block B so I will use 'M'
4.0(1.27) + M(0) = (4.0 + M)v
v = 5.1 / (4.0 + M)
4. What is the total kinetic energy of the blocks just before they hit the floor?
The initial kinetic energy just after collision will be increased by the potential energy decrease of the fall
KE = ½mv² + mgh
KE = ½(4.0 + M)(5.1 / (4.0 + M))² + (4.0 + M)9.8(0.8)
KE = 13 / (4.0 + M) + 7.84(4.0 + M)
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