Answer:
a) 16.86 m/s.
b) 15.40 m/s
c) 3.175 m/s
Explanation:
let pi be the initial momentum of the system, pf be the final momentum of the system, m1 be the mass of the bullet and V1 be the intial velocity of the bullet, m2 be the mass of the block and V2 be the intial velocity of the block
a) from the conservation of linear momentum:
pi = pf
m1×(V1)i + m2×(V2)i = Vf×(m1 + m2)
(0.0220)×(400) + (0.500)×(0) = Vf(0.0220 + 0.500)
8.8 = 0.522×Vf
Vf = 16.86 m/s
Therefore, the bullet-embedded block system will be moving with a speed of 16.86 m/s.
b) let Vf be the final velocity of the bullet-embedded block system, Wf be the work done by friction on the system, Vi be the initial velocity that the bullet-embedded block system initially moves with.
the total work done on the system is given by:
Wtot = Δk = kf - ki
Wf = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2
f×ΔxΔcos(Ф) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2
m×Δx×g×μ×(-1) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2
m×g×Δx×μ = 1/2(Vi)^2 - 1/2(Vf)^2
1/2(0.0220 + 0.500)(Vf)^2 = 1/2(0.0220 + 0.500)(16.86)^2 - (0.0220 + 0.500)×(9.8)×(8)×(0.30)
(0.261)(Vf)^2 = 86.469
(Vf)^2 = 237.2196
Vf = 15.40 m/s
Therefore, bullet-embedded block system will have a speed of 15.40 m/s after sliding of the friction surface.
c) let Vf be the speed of the bullet-embedded before hitting the block, Vi be the initial velocity of the block and V be the velocity of the bullet-embedded block and block system.
M×Vf + m×Vi = (m + M)×V
(0.0200 + 0.50)×(15.40) + (2)×(0) = (0.022 + 0.50 + 2)×V
8.008 = 2.522×V
V = 3.175 m/s
Therefore, the bullet-embedded block and block system will be moving at a speed of 3.175 m/s.
