Answer:
(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
