Question

A 0.0210 kg bullet moving horizontally at 500 m/s embeds itself into an initially stationary 0.500 kg block.

(a) What is their velocity just after the collision?
_____ m/s
(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity?
_____ m/s
(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping?
____ m

Answer 1

m_b = mass of bullet = 0.021 kg

V_ib = initial velocity of bullet before collision = 500 m/s

V_iB= initial velocity of block = 0 m/s

M_B = mass of Block = 0.5 kg

V = combined velocity after collision

a)

Using conservation of momentum

m_b V_ib + M_B V_iB = (m_b + M_B) V

0.021 (500) + (0.5) (0) = (0.021 + 0.5) V

V = 20.2 m/s

b)

retardation caused due to friction is given as = a = - g = - 0.30 x 9.8 = - 2.94 m/s²

distance travelled = d = 8 m

V_i = initial velocity just after collision = V = 20.2 m/s

V_f = final velocity

Using the kinematics equation

V_f² = V_i² + 2 a d

V_f² = 20.2² + 2 (-2.94) (8)

V_f = 19 m/s

c)

Using conservation of momentum

(m_b + M_B) V_f + M'_B V'_iB = (m_b + M_B + M'_B) V'_f

(0.021 + 0.5) (19) + 2 (0) = (0.021 + 0.5 + 2) V'_f

V'_f= 3.93 m/s

c)

retardation caused due to friction is given as = a = - g = - 0.30 x 9.8 = - 2.94 m/s²

distance travelled = d

V_i = initial velocity just after collision = V = 3.93 m/s

V_f = final velocity =0

Using the kinematics equation

V_f² = V_i² + 2 a d

0² = 3.93² + 2 (-2.94) d

d = 2.63 m