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

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.