Question

As seen in the figure, a bullet with mass of 15.0-g is fired vertically and penetrates a block with mass of 2.5-kg and the block moves upward 5.00 mm. The bullet comes to rest in the block in a time interval of 0.0011 s, and suppose that the force on the bullet is constant during this time interval. Find the kinetic energy of the bullet before penetration to the block.

Answer 1

Answer:

KE = 2.03 J

Explanation:

After impact, the kinetic energy of the bullet+block will convert to potential energy

½mv² = mgh

v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s

conservation of momentum during the collision

0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)

u = 16.4481 m/s

KE = ½mv² = ½(0.015)16.4481² = 2.0290499...

KE = 2.03 J