A 1.00 kg block is attached to a horizontal spring with spring constant 2500 N/m. The block is at rest on a frictionless surface. A 20 g bullet is fired into the block, in the face opposite to the spring and sticks.

What is the bullet's speed if the subsequent oscillations have an amplitude of 10.0 cm?

(Express your answer in m/s as a pure number).

The speed of the bullet is = 27.36

Explanation:

Mass of the block (M) = 1 kg

Spring constant = 2500 N per meter

Mass of the bullet () = 20 gm = 0.2 kg

Amplitude A = 0.1 m

Now the final velocity of block when bullet is entered into the block is given by

4.56

From conservation of momentum principal

Since

1.2 × 4.56 = 0.2

= 27.36

Therefore the speed of the bullet is = 27.36

Height of baby carriage from ground = 21m

Mass of carriage with baby = 1.5 kg

The carriage has potential energy by virtue of its height.

Potential energy = mgh = 1.5×10×21 = 315 J

Hence, potential energy of the carriage is 315 Joule.