I am studying for Finals and I am having trouble with 3 questions!! Thank you to

Question

I am studying for Finals and I am having trouble with 3 questions!! Thank you to everyone/anyone who decides to help me!!

1) An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block.   The block moves into a spring and compresses it by 4.8 cm. The force constant of the spring is 2300 N/m. In the figure, the initial velocity of the bullet is closest to:

2) Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of 0.01 rad/s2 about the center O. In the figure, the linear speed of P, when it reaches the y-axis, is closest to:

3) A potter's wheel, with rotational inertia 31 kg?m2, is spinning freely at 40.0 rpm. The potter drops a lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?

An 8 g bullet is shot into a 4.0 kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 4.8 cm. The force constant of the spring is 2300 N/m. In the figure, the initial velocity of the bullet is closest to: Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of 0.01 rad/s2 about the center O. In the figure, the linear speed of P, when it reaches the y-axis, is closest to: A potter's wheel, with rotational inertia 31 kg?m2, is spinning freely at 40.0 rpm. The potter drops a lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?

Explanation / Answer

1)E in = E out

KE= Work done by spring
.5mv^2= .5kx^2 <-----the .5's cancel
4.008v^2= 2300(.048^2)
v of block and bullet= 1.1498m/s

momentum in = momentum out
mv=(m + m2) v
.008v=(.008 + 4)(1.1498)
v=576.0498 m/s

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