A 10 0-g bullet is fired into a stationary block of wood having mass m 55.00 kg.

Question

A 10 0-g bullet is fired into a stationary block of wood having mass m 55.00 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet? A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 30.08 with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision. Two blocks are free to slide along the frictionless. Wooden track shown below. The block of mass m1 = 5.00 kg is released from the position shown, at height h = 5.00 m above the flat part of the track Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which ml rises after the elastic collision.

Explanation / Answer

4)

mb = mass of bullet = 0.01 kg

m = mass of block = 5.5 kg

Vb = velocity of bullet

V = velocity of the combination = 0.6 m/s

using conservation of momentum

mb Vb = (mb + m) V

(0.01) Vb = (0.01 + 5.5) (0.6)

Vb = 331 m/s

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