A thin block of soft wood with a mass of 0.080 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 599 m/s at a block of wood and passes completely through it. The speed of the block is 22 m/s immediately after the bullet exits the block.

(a) Determine the speed of the bullet as it exits the block. (m/s)

Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

KEi = J

KEf = J

(a) 222.126 m/s.

(b) KEi = 837.8 J, KEf = 308.81 J

Explanation:

(a)

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = mv+m'v' equation 1

Where m = mass of block, m' = mass of bullet, u = initial velocity of block, u' = initial velocity of bullet, v = final velocity of block, v' = final velocity of bullet.

make v' the subject of the equation,

v' = [(mu+m'u')-mv]/m' Equation 2

Given: m = 0.08 kg, m' = 4.67 g = 0.00467 kg, u = 0 m/s (at rest), u' = 599 m/s, v = 22 m/s.

Substitute into equation 2

v' = [(0.08×0)+(0.00467×599)-(0.08×22)]/0.00467

v' = (2.79733-1.76)/0.00467

v' = 1.03733/0.00467

v' = 222.126 m/s.

Hence the speed at which the bullet exit the block = 222.126 m/s.

(b)

Initial kinetic energy of the system

KEi = 1/2mu² + 1/2m'u'² = 1/2(0.08×0²) + 1/2(0.00467×599²)

KEi = 0+837.8 = 837.8 J.

Final kinetic energy of the system

KEf = 1/2mv² + 1/2m'v'²

KEf = 1/2(0.08×22²) + 1/2(0.00467×222.126²)

KEf = 19.36+115.21 = 308.81 J

answer: i don't know the answer good luck

explanation:

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles