Question

The 20 g bullet is traveling at 400 m/s when it becomes embedded in the 2 -kg stationary block. Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the block and the plane is μk=0.2 .

Answer 1

Answer:

The block+bullet system moves 4 m before being stopped by the frictional force.

Explanation:

Using the law of conservation of llinear momentum and the work energy theorem, we can obtain this.

According to Newton's second law of motion

Momentum before collision = Momentum after collision

Momentum before collision = (0.02×400) + 0 (stationary block)

Momentum before collision = 8 kgm/s

Momentum after collision = (2+0.02)v

8 = 2.02v

v = 3.96 m/s.

According to the work-energy theorem,

The kinetic energy of the block+bullet system = work done by Friction to stop the motion of the block+bullet system

Kinetic energy = (1/2)(2.02)(3.96²) = 15.84 J

Work done by the frictional force = F × (distance moved by the force)

F = μmg = 0.2(2.02)(9.8) = 3.96 N

3.96d = 15.84

d = (15.84/3.96) = 4 m