Answer:
weight of the bullet is 0.0500 lb
velocity as it travels from block A to block B is 900 ft/s
Explanation:
given data
horizontal velocity = 1500 ft/s
mass block A = 6-lb
mass block B = 4.95 lb
blocks A velocity = 5 ft/s
blocks B velocity = 9 ft/s
solution
we apply here law of conservation of momentum that is
m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2 ................1
put here value and we get m
m = [tex]\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}[/tex] ..............2
m = [tex]\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}[/tex]
m = 0.0500 lb
and
when here bullet is pass through the A block then moment is conserve that is
m × v(o) + m × v1 + m1 × v1 ............3
v1 = [tex]\frac{m\times v(o)- m1\times v1}{m}[/tex]
v1 = [tex]\frac{0.0500\times 1500 - 61\times 5}{0.05}[/tex]
v1 = 900 ft/s
